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TSP Java 实现

luis — Sun, 31 Mar 2013 06:39:00 GMT

package BotClean;

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution2 {

    static class Status{
        int id;//current city
        String action;
        int distance;
        Vector unvisited;
        HashSet visited;
        public Status(int id)
        {
            this.id=id;
        }
    }

    static class DirtyDot{
        int id;//current city
        int x;
        int y;

        public DirtyDot(int id, int x, int y)
        {
            this.id=id;
            this.x=x;
            this.y=y;
        }
    }
/* Head ends here */
    static void next_move(int x, int y, String[] board){
        TSP(board,x,y);

    }

/* Tail starts here */
    public static void main(String[] args) {
        //TSP(null,0,0);

        Scanner in = new Scanner(System.in);
        int [] pos = new int[2];
        String board[] = new String[5];
        for(int i=0;i<2;i++) pos[i] = in.nextInt();
        for(int i=0;i<5;i++) board[i] = in.next();
        System.out.println("begin now!");
        next_move(pos[0], pos[1], board);
    }



    public static void TSP(String[] board, int x, int y) {

        String dirty="d";
        HashMap dirtyList=new HashMap();
        int id=1;
        DirtyDot tempDirtyDot2;
        DirtyDot tempDirtyDot=new DirtyDot(id-1,x,y);
        dirtyList.put(id,tempDirtyDot);

        for(int i=0;i        {
            if(board[i].contains(dirty))
            {
                int position;
                position=board[i].indexOf(dirty, 0);
                while(position!=-1)
                {
                    tempDirtyDot=new DirtyDot(id,i,position);
                    dirtyList.put(id,tempDirtyDot);
                    id++;
                    position=board[i].indexOf(dirty,position+1);
                    System.out.println("position"+position);
                }
            }
        }

//        int MAX=id;
//        int dist[][]=new int[MAX][MAX];
//
//        for(int i=0;i//            for(int j=0;j//               {
//                tempDirtyDot=dirtyList.get(i);
//                tempDirtyDot2=dirtyList.get(j);
//                 dist[i][j]=Math.abs(tempDirtyDot.x-tempDirtyDot2.x)+Math.abs(tempDirtyDot.y-tempDirtyDot2.y);
//               }

          int MAX=4;

          int dist[][]={{-1,3,6,7},{2,-1,8,6},{7,3,-1,5,},{7,3,7,-1}};

        Vector currentStatus=new Vector();
        Vector previousStatus;
        /*Initialize current Status Vector*/
        for(int i=1;i        {
            Status status=new Status(i);
            status.distance=dist[i][0];
            status.unvisited=new Vector();
            status.visited=new HashSet();
            currentStatus.add(status);
        }
        System.out.println("Initialized current Status Vector");
        //System.out.println(currentStatus.size());

        for(int j=0;j        {
            previousStatus=currentStatus;
            currentStatus=new  Vector();
            //System.out.println(previousStatus.size());
            for(int i=1;i// enumerate each node
            {
                for(int k=0;k                {
                    Status tempStatus=previousStatus.elementAt(k);
                    if(isContain(tempStatus,i)==false)//gurantee node i is not in tempStatus
                    {
                        Status newStatus=new Status(i);
                        newStatus.distance=tempStatus.distance+dist[i][tempStatus.id];
                        System.out.println("id"+tempStatus.id);
                        newStatus.visited=new HashSet(tempStatus.visited);
                        newStatus.unvisited=new Vector(tempStatus.unvisited);
                        newStatus.unvisited.add(0,(Integer)(tempStatus.id));
                        newStatus.visited.add(tempStatus.id);
                        currentStatus.add(newStatus);
                        System.out.println(newStatus.unvisited.size());
                        System.out.println(newStatus.distance);

                    }
                }
            }
            //System.out.println(currentStatus.size());
            currentStatus=optimize(currentStatus);//dp process
            //System.out.println(currentStatus.size());
        }//end for

        Iterator iterator = currentStatus.iterator();

        Status tempStatus=iterator.next();
        Status shortest=tempStatus;
        int minDistance=dist[0][tempStatus.id]+tempStatus.distance;
        System.out.println("1:"+tempStatus.distance);
        System.out.println("2:"+dist[0][tempStatus.id]);

        while (iterator.hasNext()) {
             tempStatus=iterator.next();
             int tempDistant=dist[0][tempStatus.id]+tempStatus.distance;
             System.out.println("11:"+tempStatus.distance);
             System.out.println("22:"+dist[0][tempStatus.id]);
             System.out.println("33:"+tempStatus.id);

             if(tempDistant             {
                 minDistance=tempDistant;
                 System.out.println("in loop"+minDistance);
                 //System.out.println(tempStatus.distance);
                 shortest=tempStatus;
             }
          }

        System.out.println("distance: "+minDistance);
        System.out.println("size:"+shortest.unvisited.size());
        System.out.print(" 1");
        System.out.print(" "+shortest.id);
        for(int i=0;i            {
                 int tmp=shortest.unvisited.get(i)+1;
               System.out.print(" "+tmp);
            }
        System.out.println(" 1");
    }

    private static Vector optimize(Vector cs) {
        Status tempStatus,anotherStatus;
        int j;

        Iterator iterator = cs.iterator();
         while (iterator.hasNext()) {
             tempStatus=iterator.next();
              for(j=0;j                {
                      anotherStatus=cs.get(j);
                      if(tempStatus.id==anotherStatus.id&&tempStatus.visited.equals(anotherStatus.visited))
                      {
                          if(tempStatus.distance>anotherStatus.distance)
                          {
                              iterator.remove();
                             break;
                          }
                      }
                }

         }
          return cs;

    }

    static boolean isContain(Status sta, int i)
    {
        if(i==sta.id)
            return true;
        else
          return sta.unvisited.contains(i);
     }


}

luis 2013-03-31 14:39 发表评论

动态规�?解TSP 旅行商问�?�l�典实现

luis — Sat, 30 Mar 2013 21:26:00 GMT

原文地址: http://blog.csdn.net/gfaiswl/article/details/4749713

1.问题定义

TSP问题�Q�旅行商问题�Q�是指旅行家要旅行n个城市，要求各个城市�l�历且仅�l�历一�ơ然后回到出发城市，�q�要求所走的路程最短�?/p>

假设现在有四个城市，0,1,2,3�Q�他们之间的代�h如图一�Q�可以存成二�l�表的�Ş�?br />

图一

现在要从城市0出发�Q�最后又回到0�Q�期�?�Q?�Q?都必��dƈ且只能经�q�一�ơ，使代��h��?/p>

2.动态规划可行�?/h1>
设s, s1, s2, …, sp, s是从s出发的一条�\径长度最短的��单回路，假设从s��C��一个城市s1已经求出�Q�则问题转化为求从s1到s的最短�\径，昄��s1, s2, …, sp, s一定构成一条从s1到s的最短�\径，所以TSP问题是构成最优子�l�构性质的，用动态规划来求解也是合理的�?/p>

3.推导动态规划方�E?/h1>
        假设从顶点s出发�Q��od(i, V’)表示从顶点i出发�l�过V’(是一个点的集�?中各个顶点一�ơ且仅一�ơ，最后回到出发点s的最短�\径长度�?/p>
        推导�Q?分情冉|��讨论)
        ①当V’为空集，那么d(i, V’)�Q�表�C�Z��i不经�q��Q何点��回到s了，如上囄�� 城市3->城市0(0��v点城�?。此时d(i, V’)=Cis(��是城市i �?城市s 的距��?�?/p>
        ②如果V’不�ؓ�I�，那么��是对子问题的最优求解。你必须在V’�q�个城市集合中，��试每一个，�q�求出最优解�?/p>
           d(i, V’)=min{Cik + d(k, V’-{k})}
           注：Cik表示你选择的城市和城市i的距��，d(k, V’-{k})是一个子问题�?/p>
        �l�g��所�q�ͼ�TSP问题的动态规划方�E�就出来了：

4.实例分析

现在寚w��题定义中的例子来说明TSP的求解过�E��?假设出发城市�?0城市)

①我们要求的最�l�结果是d(0,{1,2,3}),它表�C�，从城�?开始，�l�过{1,2,3}之中的城市�ƈ且只有一�ơ，求出最短�\�?

②d(0,{1,2,3})是不能一下子求出来的�Q�那么他的值是怎么得出的呢�Q�看上图的第二层�Q�第二层表明了d(0,{1,2,3})所需依赖的倹{��那么得出：

d(0,{1,2,3})=min {

C01+d(1,{2,3})

C02+d{2,{1,3}}

C03+d{3,{1,2}}

}

③d(1,{2,3})�Q�d(2,{1,3})�Q�d(3,{1,2})同样也不是一步就能求出来的，它们的解一样需要有依赖�Q�就比如说d(1,{2,3})

d(1,{2,3})=min{

C12+d(2,{3})

C13+d(3,{2})

}

d(2,{1,3})�Q�d(3,{1,2})同样需要这么求�?/p>

④按照上面的思�\�Q�只有最后一层的�Q�当当V’为空集时�Q�Cis的值才可以求，它的值是直接�?/p>

�q�张表里求得的�?/p>

5.�~�程思�\

��d(i, V’)转换成二�l�表�Q�d[i][j]

在程序中模拟填表的过�E�，主要要考虑到j�q�个参数的表�C�，它要代表一个集合，可以用二�l�数�l�来表示�?/p>

   6.源代�?/h1>
注：�׃��本�h水��^有限�Q��ƈ且主要在�q�里是体现思�\�Q�所以程序�ƈ不是很完善，代码质量也不高，很地方可以写得通用一些，所以这里只是提供一个参考，�E�序的进一步完善，��p��者自由发挥�?/p>
#include
#include
int IsIncluded(int x,int array[3])//x是否包含在数�l�中
{
    if((array[0] != x) && (array[1] != x) && (array[2] != x))
        return 0;
    return 1;
}
int Left(int k,int array[3],int V[8][3])//实现V'-{k} 的下标检�?nbsp;
{
    int i = 0,index = 0,array_0_count = 0,array_1_count = 0,array_2_count = 0,array_3_count = 0;
    int V_0_count = 0,V_1_count = 0,V_2_count = 0,V_3_count = 0;
    int temp[3];
    for(i = 0; i < 3; i++)
        temp[i] = array[i];
    for(i = 0; i < 3; i++)
        if(temp[i] == k)
            temp[i] = 0; //相当于去掉k�q�个城市
    for(i = 0; i < 3; i++)
    {
        if(temp[i] == 0)
            array_0_count++;
        else if(temp[i] == 1)
            array_1_count++;
        else if(temp[i] == 2)
            array_2_count++;
        else
            array_3_count++;
    }
    for(index = 0; index < 8; index++)
    {
        for(i=0; i < 3; i++)
        {
            if(V[index][i] == 0)
                V_0_count++;
            else if(V[index][i] == 1)
                V_1_count++;
            else if(V[index][i] == 2)
                V_2_count++;
            else
                V_3_count++;
        }
        if((array_0_count == V_0_count) && (array_1_count == V_1_count)
            && (array_2_count == V_2_count) && (array_3_count == V_3_count))
            return index;
        V_0_count = 0;
        V_1_count = 0;
        V_2_count = 0;
        V_3_count = 0;
    }
    return 0;
}
void TSP(int d[4][8],int c[4][4],int V[8][3],int n)
{
    int i = 0,j = 0,k = 0;
    for(i = 1; i < n; i++)//V'为空�Ӟ��l�赋��|��
        d[i][0] = c[i][0];
    for(j = 1; j < 7; j++)//按列遍历不同集合�Q�{1},{2},{3},{1,2},{1,3}.....
    {
        for(i = 1; i < n; i++)//遍历城市1�Q?�Q?
        {
            if( !IsIncluded(i,V[j]) )//i必须不在集合中，否则��属于经�q�两�ơ，不符合题�?nbsp;
            {
                for(k = 0; k < 3; k++)//分别试探集合中的每一点，取最��?nbsp;
                {
                    if((V[j][k] != 0) && ((c[i][V[j][k]] + d[V[j][k]][Left(V[j][k],V[j],V)]) < d[i][j]))
                        d[i][j] = c[i][V[j][k]] + d[V[j][k]][Left(V[j][k],V[j],V)];
                }
            }
        }//end of     for(i = 1; i < n; i++)//遍历城市1�Q?�Q?
    }//end of for(j = 1; j < ((int)pow(2,n)-1); j++)
    for(k = 0; k < 3; k++)//分别试探下一步�ؓ集合中的��M��一点，取最��?nbsp;
    {
        if((V[7][k] != 0) && (c[0][V[7][k]] + d[V[7][k]][Left(V[7][k],V[7],V)]) < d[0][7])
            d[0][7] = c[0][V[7][k]] + d[V[7][k]][Left(V[7][k],V[7],V)];
    }
}
void main()
{
    int V[8][3]=
    {
        0,0,0,
        0,0,1,
        0,0,2,
        0,0,3,
        0,1,2,
        0,1,3,
        0,2,3,
        1,2,3
    };
    int c[4][4]=
    {
        0,3,6,7,
        5,0,2,3,
        6,4,0,2,
        3,7,5,0
    };
    int d[4][8]={0},i=0,j=0;
    for(i=0; i<4; i++)
        for(j=0; j<8; j++)
            d[i][j]=1000;   //假设1000为无�I�大
    TSP(d,c,V,4);
    printf("The least road is:%d/n",d[0][7]);
}

luis 2013-03-31 05:26 发表评论

luis — Wed, 05 Dec 2012 10:44:00 GMT

相当于A(B x)

luis 2012-12-05 18:44 发表评论

Python extract all comments:提取所有comments,提取c/c++中注释Python脚本

luis — Mon, 03 Dec 2012 00:35:00 GMT

注意�Q�我们只是简单的提取// 以及/* */之间的内宏V�?br />如果 �E�序中出��C��“/*”,会有bug

#!/usr/bin/env python

import sys
import re

def comment_finder(text):
    pattern = re.compile( r'//.*?$|/\*.*?\*/', re.DOTALL | re.MULTILINE)
    result = pattern.findall(text)
    return result

def print_command(filename):

    codefile = open(filename,'r')
    commentfile = open(filename+".txt",'w')
    lines=codefile.read()
    codefile.close()
    #the list of comments
    list_of_comments = comment_finder(lines)
    for comment in list_of_comments:
        #print comment[0:2]
        if comment[0:2] == "//":
                comment_to_write = comment[2:]
        else:
            comment_to_write = comment[2:-2]
        if len(comment_to_write)!=0:
            commentfile.write(comment_to_write)
        commentfile.write('\n')
    commentfile.close()

if __name__ == "__main__":
    for filename in sys.argv[1:]:
        print_command(filename)

使用�Q?br />

在linux下面转到当前目录 ./get_comment.py *
或�?指定文�g�c�d��

./get_comment.py *.c

luis 2012-12-03 08:35 发表评论

Latex 表格

luis — Wed, 21 Nov 2012 17:07:00 GMT

\begin{tabular}{|c|c|c|c|c|c|}
\hline
Accuracy & General & test 1 & test 2 & test 3 & test 4\tabularnewline
\hline

Linear  & 0.8950695  & 0.8817864 & 0.8884132  & 0.8884036  & 0.9217801
\tabularnewline
\hline

Polynomial Degree=2 & 0.927821  & 0.9118337 & 0.934325 & 0.913003 &
0.9335226\tabularnewline
\hline
Polynomial Degree=2 & 0.8350725  & 0.8317865 & 0.8285052  & 0.8278306  &
0.852091
\tabularnewline
\hline
Polynomial Degree=3 & 0.770894 & 0.7539803  & 0.7510282 &  0.7762011 &
0.7821553\tabularnewline
\hline
Polynomial Degree=4 & 0.7172013 & 0.685079 & 0.6988003 & 0.7329987 &
0.7520278\tabularnewline

\hline
\end{tabular}

=========================

Type one:

%%%%%%%%%%��单表�?%%%%%%%%%

\begin{tabular}{|c|c|}

\hline

a & b \\\hline

c & d\\

\hline

\end{tabular}

Type two:

%%%%%%%%%%��单表�?居中(一)%%%%%%%%%%

\begin{center}

\begin{tabular}{|c|c|}

\hline

a & b \\\hline

c & d\\

\hline

\end{tabular}

\end{center}

Type three:

%%%%%%%%%%��单表�?居中�Q�二�Q?%%%%%%%%%

\centering

\begin{tabular}{|c|c|}

\hline

a & b \\\hline

c & d\\

\hline

\end{tabular}

This a Table~??

Type four:

%%%%%%%%%%��单表�?居中,标题�Q�编�?固定位置%%%%%%%%%%

\begin{table}[h]

\centering

\caption{table}\label{tab:table}

\begin{tabular}{|c|c|}

\hline

a & b \\\hline

c & d\\

\hline

\end{tabular}

\end{table}

Type five:

%%%%%%%%%%一个完整的例子%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%Beginning of Table example%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{table}[ht]

\caption{Review paper} % title of Table

\centering % used for centering table

\begin{tabular}{|c|c|} % centered columns (2 columns)

\hline\hline %inserts double horizontal lines

Item 1 & Item 2 \\ [0.5ex] % inserts table heading

\hline % inserts single horizontal line

a & b \\ % inserting body of the table

c & d \\

e & f \\[1.5ex] % [1.5ex] adds vertical space

\hline %inserts single line

\end{tabular}

\label{table:nonlin} % is used to refer this table in the text

\end{table}

%%%%%%%%%%%%%%%%%% end of the example%%%%%%%%%%%%%%%%%%%%%%%%%

luis 2012-11-22 01:07 发表评论

luis — Wed, 21 Nov 2012 16:44:00 GMT

原文�Q?a >http://www.cnblogs.com/tonyqus/archive/2004/12/07/73710.html

熟悉C++的�h对于两个字符串比较的代码一定很了解�Q?nbsp;
(string1==string2)
但在java中，�q�个代码即��在两个字�W�串完全相同的情况下也会�q�回false
Java中必��M��用string1.equals(string2)来进行判�?nbsp;

补充
如果�Q?nbsp;
string s1=new String("Hello");
string s2=new String("Hello");
�?s1==s2)=false

如果�Q?nbsp;
string s1="Hello";
string s2="Hello";
�?s1==s2)=true;
因�ؓ他们指向的同一个对象�?nbsp;

如果把其他变量的��D��l�s1和s2�Q�即使内容相同，�׃��不是指向同一个对象，也会�q�回false。所以徏议��用equals()�Q�因为equals比较的才是真正的内容

luis 2012-11-22 00:44 发表评论

Latex 多个囄��

luis — Thu, 15 Nov 2012 19:43:00 GMT

\usepackage{subfigure}

注意不用�?{subfig}
\begin{figure}[ht]

\centering

\subfigure[$\alpha$=0.5,$\gamma$=0.9,$\epsilon$=0.01]{

\includegraphics[width=4.4in] {2.png}

}

\subfigure[$\alpha$=0.7,$\gamma$=0.9,$\epsilon$=0.01]{

\includegraphics[width=4.4in] {3.png}

}

\subfigure[$\alpha$=0.9,$\gamma$=0.9,$\epsilon$=0.01]{

\includegraphics[width=4.4in] {4.png}

}

\subfigure[$\alpha$=0.7,$\gamma$=0.9,$\epsilon$=0.1]{

\includegraphics[width=4.4in] {5.png}

}

\subfigure[$\alpha$=0.7,$\gamma$=0.9,$\epsilon$=0.2]{

\includegraphics[width=4.4in] {6.png}

}

\subfigure[$\alpha$=0.7,$\gamma$=0.6,$\epsilon$=0.01]{

\includegraphics[width=4.4in] {1.png}

}

\end{figure}

luis 2012-11-16 03:43 发表评论

luis — Thu, 08 Nov 2012 00:21:00 GMT

import math
def mianji(n,s):

temp=1/4*n*(s**2)/math.tan(math.pi/n)

return temp

print mianji(5,7)
============
使用时math.pi math.tan

luis 2012-11-08 08:21 发表评论

luis — Wed, 07 Nov 2012 22:56:00 GMT

import random

a=random.randint(0,1001)#�q�个生成的区间是 [0,1001]
a=random.randrange(0,10001)#生成的是区间是[0,1001)

print a
random.randrange([start], stop[, step])
Return a randomly selected element from range(start, stop, step). This is equivalent tochoice(range(start, stop, step)), but doesn’t actually build a range object.

And range(start, stop) returns [start, start+step, ..., stop-1], not [start, start+step, ..., stop]. As for why... zero-based counting rules and range(n) should return n elements, I suppose. Most useful for getting a random index, I suppose.

While randint is documented as:

random.randint(a, b)
Return a random integer N such that a <= N <= b. Alias for randrange(a, b+1)

So randint is for when you have the maximum and minimum value for the random number you want.

luis 2012-11-08 06:56 发表评论

Python �W�记2 //

luis — Wed, 07 Nov 2012 22:43:00 GMT

Python “” ‘’ 都可以表�C�字�W�串

s1 = "hello,world"
如果要写成多行，那么��p��使用/ (“�q�行�W?#8221;)吧，�?nbsp;
s2 = "hello,/
world"
s2与s1是一��L��。如果你�?个双引号的话�Q�就可以直接写了�Q�如下：
s3 = """hello,
world,
hahaha."""�Q�那么s3实际上就�?hello,/nworld,/nhahaha.", 注意“/n”

s5 = "Let's go"
s4 = 'Let/'s go'

我们也可以把''' ''' 作�ؓ多行注释

str(object) 可以��所有�{化�ؓ字符丌Ӏ?br />

python	java	描述
or	\|\|	逻辑�?/td>
and	&&	逻辑�?/td>
not	�Q?/td>	逻辑�?/td>
<�Q?gt;�Q?lt;=�Q?gt;=�Q?=�Q?=�?lt;>	<�Q?gt;�Q?lt;=�Q?gt;=�Q?=�Q?=	比较操作
is�Q�is not	instanceof	�w�䆾认证
\|	\|	位或
&	&	位与
^	^	位异�?/td>
<<�Q?gt;>	<<�Q?gt;>	�U�M��
+�Q?�Q?�Q?	+�Q?�Q?�Q?	加减乘除
%	%	余数
~	~	位取�?/td>

//�q�算�W?
10/3==3
120//10==12
121//10==12
122//10==12
130//10==13
10//3.0==3.0

A new operator, //, is the floor division operator. (Yes, we know it
looks like C++'s comment symbol.) // always performs floor division no
matter what the types of its operands are, so 1 // 2 is 0 and 1.0 //
2.0 is also 0.0.

not ()

luis 2012-11-08 06:43 发表评论

Latex 多行公式

luis — Mon, 05 Nov 2012 22:29:00 GMT

\newcommand{\parenthnewln}[1]{\right.\\#1&\left.{}}
\begin{equation}
\begin{split}
f(x)=1+&g(x)\\
=1+&\left(x+x^2+\dots
\parenthnewln{+}x^n+\ldots\right)
\end{split}
\end{equation}

得到�Q?br />

luis 2012-11-06 06:29 发表评论

luis — Wed, 31 Oct 2012 13:30:00 GMT

http://en.wikibooks.org/wiki/LaTeX/Algorithms_and_Pseudocode
Typesetting using the `algorithm2e` package

The algorithm2e package (first released 1995, latest updated December 2009 according to the v4.01 manual) allows typesetting algorithms with a lot of customization. The package is loaded like

\usepackage[options]{algorithm2e} 

and a simple example, taken from the v4.01 manual, is

\begin{algorithm}[H]
 \SetAlgoLined
 \KwData{this text}
 \KwResult{how to write algorithm with \LaTeX2e }
 initialization\;
 \While{not at end of this document}{
  read current\;
  \eIf{understand}{
   go to next section\;
   current section becomes this one\;
   }{
   go back to the beginning of current section\;
  }
 }
 \caption{How to write algorithms}
\end{algorithm}

which produces

luis 2012-10-31 21:30 发表评论

luis — Wed, 31 Oct 2012 13:25:00 GMT

\usepackage{listings}
\usepackage{color}

\definecolor{dkgreen}{rgb}{0,0.6,0}
\definecolor{gray}{rgb}{0.5,0.5,0.5}
\definecolor{mauve}{rgb}{0.58,0,0.82}

\lstset{frame=tb,
  language=Java,
  aboveskip=3mm,
  belowskip=3mm,
  showstringspaces=false,
  columns=flexible,
  basicstyle={\small\ttfamily},
  numbers=none,
  numberstyle=\tiny\color{gray},
  keywordstyle=\color{blue},
  commentstyle=\color{dkgreen},
  stringstyle=\color{mauve},
  breaklines=true,
  breakatwhitespace=true
  tabsize=3
}
================

You can change default language in the middle of document with \lstset{language=Java}.

Example of usage in the document:

\begin{lstlisting}
// Hello.java
import javax.swing.JApplet;
import java.awt.Graphics;

public class Hello extends JApplet {
    public void paintComponent(Graphics g) {
        g.drawString("Hello, world!", 65, 95);
    }
}

\end{lstlisting}

Here's the result:

luis 2012-10-31 21:25 发表评论

label switching

luis — Tue, 30 Oct 2012 19:36:00 GMT

label switching:
比如有两��p��果： �W�一个箱子内有两个苹果，label 为a 的概率�ؓ30%,为b的概�?0%�Q�第二个��子内有四个�Ҏ��Q�label为b的概�?0%,label 为a的概�?0%.
如果我们求所有的�Ҏ��的重量，只需要将所有的��子内的�Ҏ��取出来求重量卛_��?br />但是我们先求label a的箱子苹果的重量�Q�加上label b的箱子苹果的重量�Q�可能出��C��ơ取的是同一个箱子，�q�就是label switching问题�?img src ="http://www.shnenglu.com/luyulaile/aggbug/194109.html" width = "1" height = "1" />

luis 2012-10-31 03:36 发表评论

Python generate corpus using Dirichlet distribution

luis — Sun, 28 Oct 2012 02:13:00 GMT

At first, let's define the sample function:

def sample(dist, num_samples=1):
    """
    Uses the inverse CDF method to return samples drawn from an
    (unnormalized) discrete distribution.

    Arguments:

    dist -- (unnormalized) distribution

    Keyword arguments:

    num_samples -- number of samples to draw
    """

    cdf = cumsum(dist)
    r = uniform(size=num_samples) * cdf[-1]

    return cdf.searchsorted(r)

As we can see, the sample function input two parameters, one is dist, which can be an un-normalized distribution, another is the sample we want to draw.

Let's see how to generate corpus for Dirichlet--multinomial unigram language model

def generate_corpus(beta, mean, N):
    """
    Returns a corpus of tokens drawn from a Dirichlet--multinomial
    unigram language model. Each token is an instance of one of V
    unique word types, represented by indices 0, , V - 1.

    Arguments:

    beta -- concentration parameter for the Dirichlet prior
    mean -- V-dimensional mean of the Dirichlet prior
    N -- number of tokens to generate
    """

    pass # YOUR CODE GOES HERE
    #print mean
    #print beta
    #print dot(mean,beta)
    #print dirichlet(mean*beta,size=1)
    temp=sample(dirichlet(beta*array(mean),size=1),N)
    #print temp
    return temp

please keep in mind the dirichlet function is “from numpy.random.mtrand import dirichlet"
and the parameters it receives are corresponding to beta*array(mean). beta is the concentration factor, and mean is the vector which sum to 1.

another way is to generate corpus is using the property:
P(D'|D,H)= Nv+beta_nv/N+beta

def generate_corpus_collapsed(beta, mean, N):
    """
    Returns a corpus of tokens drawn from a Dirichlet--multinomial
    unigram language model using the 'collapsed' generative process
    (i.e., phi is not explicitly represented). Each token is an
    instance of one of V unique word types.

    Arguments:

    beta -- concentration parameter for the Dirichlet prior
    mean -- V-dimensional mean of the Dirichlet prior
    N -- number of tokens to generate
    """

    V = len(mean) # vocabulary size

    corpus = zeros(N, dtype=int) # corpus

    Nv = zeros(V, dtype=int) # counts for each word type

    pass # YOUR CODE GOES HERE
    for n in xrange(N):
        corpus[n]=sample((Nv+beta*array(mean))/(n+beta),1)
        Nv[corpus[n]]+=1;
    return corpus

Let's see how to generate corpus for Mixture of Dirichlet-multinomial unigram language model

def generate_corpus(alpha, m, beta, n, D, Nd):
    """
    Returns a grouped corpus drawn from a mixture of
    Dirichlet--multinomial unigram language models.

    Arguments:

    alpha -- concentration parameter for the Dirichlet prior over theta
    m -- T-dimensional mean of the Dirichlet prior over theta
    beta -- concentration parameter for the Dirichlet prior over phis
    n -- V-dimensional mean of the Dirichlet prior over phis
    D -- number of documents to generate
    Nd -- number of tokens to generate per document
    """
    corpus = GroupedCorpus()

    pass # YOUR CODE GOES HERE
    #determine the topic the distribution for topic dirichlet(dot(m,alpha),size=1)
    #given the topic, the distribtuion for word dirichlet(dot(n,beta),size=1)
    theta=dirichlet(alpha*array(m),1)
    phis=dirichlet(beta*array(n),len(m))
    for d in range(0,D):
        [t]=sample(theta,1)
        #print groupVcab
        corpus.add(str(d),str(t),[str(x) for x in sample(phis[t,:],Nd)])
    return corpus

注意是T个topic (group)�Q?span style="font-size: 13px; background-color: #eeeeee; "> phis=dirichlet(beta*array(n),len(m)) 产生了T�?dirichlet distribution,相同的topic t应该取同一�?dirichlet distribution phis[t,:]

luis 2012-10-28 10:13 发表评论

多维矩阵求和

luis — Sat, 27 Oct 2012 05:11:00 GMT

>> test3=cat(3,test1,test2)

test3(:,:,1) =

1 2

3 4

test3(:,:,2) =

5 6

7 8

=======================================

>> sum(test3,1) %对第一个维度求�?/div>

ans(:,:,1) =

4 6

ans(:,:,2) =

12 14

>> sum(test3,2) %对第二个�l�度求和

ans(:,:,1) =

ans(:,:,2) =

>> sum(test3,3)%相当于对�W�三个维度上求和

ans =

6 8

10 12

>> sum(test3(:))

ans =

luis 2012-10-27 13:11 发表评论

matlab 矩阵合�ƈ

luis — Sat, 27 Oct 2012 04:31:00 GMT

>> test1=[1,2;3,4]

test1 =

1 2

3 4

>> test2=[5,6;7,8]

test2 =

5 6

7 8

>> test3=[test1,test2]

test3 =

1 2 5 6

3 4 7 8

>> test4=[test1;test2]

test4 =

1 2

3 4

5 6

7 8
===

对于�l�数相同的矩�?/div>

cat(1,A,B�Q�相当于[A;B]

cat (2, A, B) 相当于[A,B]

cat (3, A, B) 相当于增加维�?/div>

当A�Q�B分别��Z��l�矩阉|��Q�合�q�之后�ؓ三维矩阵�Q�A,B��Z��l�矩阵，则在�W�三�l�方向上合�ƈA,B�?/span>

>> cat(1,test1,test2)

ans =

1 2

3 4

5 6

7 8

>> cat(2,test1,test2)

ans =

1 2 5 6

3 4 7 8

>> cat(3,test1,test2)

ans(:,:,1) =

1 2

3 4

ans(:,:,2) =

5 6

7 8

luis 2012-10-27 12:31 发表评论

luis — Fri, 12 Oct 2012 01:25:00 GMT

latex 囄��l�常多出�?br />

1�U�方�? 忽略��学
[!hb]

2另一�U�方�?br />H 插入当前位置

\usepackage{float}

\begin{figure}[H]
foo
\end{figure}

看图�Q?img src ="http://www.shnenglu.com/luyulaile/aggbug/193197.html" width = "1" height = "1" />

luis 2012-10-12 09:25 发表评论

Latex 常用命��o

luis — Thu, 11 Oct 2012 21:02:00 GMT

\overline

^ _

\dfrac{}{}

\frac{}{}

\textbf

\sqrt 开根号

\tilde 斜线
\alpha \Alpha

\hat

\leq <=
\rightarrow ->
\geq >= greater equal
\begin{itemize}
\end{itemize}
\xi
\parell
求导�?\delta

关于title

\documentclass{book}

\title{TITLE}

\author{AUTHOR}

\begin{document}

\maketitle

\end{document}

关于表格

artice的文章结�?br />title
section
subsection
subsubsection

首行�~�进

\usepackage{indentfirst}

===itemlize description enumerata的区�?==

\begin{itemize}
    \item First level, itemize, first item
    \begin{itemize}
        \item Second level, itemize, first item
        \item Second level, itemize, second item
        \begin{enumerate}
            \item Third level, enumerate, first item
            \item Third level, enumerate, second item
        \end{enumerate}
    \end{itemize}
    \item First level, itemize, second item
\end{itemize}
==

主要是：

��d��Q?br />

\begin{figure}[ht]

\centering

\includegraphics[width=0.8\textwidth,clip]{image/fig2.png}

\caption{}

\label{fig:11}

\end{figure}

多个图：

\begin{figure}
  \centering
  \subfigure[Small Box with a Long Caption]{
    \label{fig:subfig:a} %% label for first subfigure
    \includegraphics[width=1.0in]{graphic.eps}}
  \hspace{1in}
  \subfigure[Big Box]{
    \label{fig:subfig:b} %% label for second subfigure
    \includegraphics[width=1.5in]{graphic.eps}}
  \caption{Two Subfigures}
  \label{fig:subfig} %% label for entire figure
\end{figure}

luis 2012-10-12 05:02 发表评论

Matlab 常用��d��公式

luis — Thu, 11 Oct 2012 20:31:00 GMT

http://www.mathworks.com/help/matlab/ref/plot.html

例子�Q?br />
�q�是丄��单的例子个大�Ӟ��?≤x≤2p区间内，�l�制曲线y1=2e-0.5x和y2=cos(4πx)�Q��ƈ�l�图形添加图形标注�?nbsp;
x=0:pi/100:2pi;
y1=2exp(-0.5x);
y2=cos(4pi*x);
plot(x,y1,x,y2)
title('x from 0 to 2{\pi}');             %加图形标�?nbsp;
xlabel('Variable X');                      %加X轴说�?nbsp;
ylabel('Variable Y');                      %加Y轴说�?nbsp;
text(0.8,1.5,'曲线y1=2e^{-0.5x}');       %在指定位�|�添加图形说�?nbsp;
text(2.5,1.1,'曲线y2=cos(4{\pi}x)');
legend('y1','y2')                         %加图�?nbsp;
复制代码

plot

2-D line plot

Syntax

plot(Y) plot(X1,Y1,...,Xn,Yn) plot(X1,Y1,LineSpec,...,Xn,Yn,LineSpec) plot(...,'PropertyName',PropertyValue,...) plot(axes_handle,...) h = plot(...) 1、设�|�图�U�宽�?/strong> set( haxis, 'LineWidth', 1.0 ); ----�q�是 set函数�Q?'LineWidth'��是axis的线宽度属性，其值默认�ؓ0.5�Q�这里可以改�?.0了�?/span>

2、调整坐标��u上下�?br />set( haxis, 'XLim', [ 2 20 ] ); set( haxis, 'YLim', [ 2 20 ] );---调整坐标轴上下限的。Z��_��那同理set( haxis, 'ZLim', [ Zmin, %% Zmax ] ) 3、调整坐标��u上的标注数字 set( haxis, 'XTick', 2:1:20 ); set( haxis, 'YTick', 2:1:20 );----是调整坐标��u上那些标注出来的数字了，2:1:20意思是�?开始，每隔1标注一�ơ，直到20为止。注意这里的2, 20最好和上面相应的XLim, YLim最大最��g��致�?/tt>

4、MATLAB中提供的�U�型及颜色属性：

如：plot(x1,y1,'r-'),表示�Q�用�U�色实线��d��囑�Ş

�U�型	说明	标记�W?/p>	说明	颜色	说明
-	实线(默认)	+	加号�W?/p>	r	�U�色
--	双划�U?/p>	o	�I�心�?/p>	g	�l�色
:	虚线	*	星号	b	蓝色
:.	点划�U?/p>	.	实心�?/p>	c	青绿�?/p>
		x	叉号�W?/p>	m	�z�红�?/p>
		s	正方�?/p>	y	黄色
		d	菱�Ş	k	黑色
		^	上三角�Ş	w	白色
		v	下三角�Ş
		>	右三角�Ş
		<	左三角�Ş
		p	五角�?/p>
		h	六边�?/p>

5、图形标题、��u标注、图形说明的讄��

囑֐�标注可用�Q�title('xx关系�?) axis([0,22,0,3]);---昄��范围为：X轴从0-22�Q?Y轴从0-3昄��?br />xlabel('电压�Q�V�Q?)�Q�ylabel('甉|��Q�A�Q?) 分别表示在X轴下标示电压�Q�V�Q�，Y轴旁标示“甉|��Q�A�Q?#8221;

legend('A曲线 ','B曲线','C曲线') 用于说明图中的曲�U�的说明�Q�顺序和plot(x1,y1,x2,y2,x3,y3)的曲�U?�?�?相同卛_��?/tt>

set(h,'LineWidth',1.5)%讄��囄��_�细

%set(gca,'XTickLabel',{'-4/T','-3/T','-2/T','-1/T','0','1/T','2/T','3/T','4/T',})

`scatter`

Scatter plot

Syntax scatter(X,Y,S,C) scatter(X,Y) scatter(X,Y,S) scatter(...,markertype) scatter(...,'filled') scatter(...,'PropertyName',propertyvalue) scatter(axes_handle,...) h = scatter(...)

luis 2012-10-12 04:31 发表评论

Matlab 常用function �W�记 luis — Thu, 11 Oct 2012 00:28:00 GMT repmat(A, 2 *3) 负责A 2*3 eye(1) [1] eye(2) 2*2的单位矩�?br />ones(n), n*n的矩�?br />ones(n,m) 或�?ones([n,m]) n*m全是1的矩�?br />zeros(n) zeros([n,m]) 全是0的矩�?br /> reshape() ��数据一列一列重新分�?br />例如�Q?br /> Reshape a 3-by-4 matrix into a 2-by-6 matrix. A = 1 4 7 10 2 5 8 11 3 6 9 12 B = reshape(A,2,6) B = 1 3 5 7 9 11 2 4 6 8 10 12 B = reshape(A,2,[]) B = 1 3 5 7 9 11 2 4 6 8 10 12 A=rand(3,3); B=A';%转置 C=inv(B);%求�?/span> D=rank(A);%求矩�늚��|?/span> [V E]=eig(A);%求矩�늚�特征向量V和特征值D 如果不是斚w��Q�可以用pinv求伪�?/span> 圆括��L��作用�Q�用于取��|��引用数组中的元素�Q?br />>> A =[1 4 7 10 2 5 8 11 3 6 9 12] A = 1 4 7 10 2 5 8 11 3 6 9 12 A(3)=3 A(10)=10 A(2,3)=5 注意不是�Q�而是逗号, A(2,:)=��L��有的�W�二�?2 5 8 11 �Ҏ��L��作用�Q?br />Matlab �Ҏ��?#8220;[ ]”的作�?br /> 1定义矩阵�Q�向量，数组�Q��?br /> 例如 a=[1 2 3;4 5 6] 2.定义函数的返回参数列表（输出变量宗量�Q��?br /> 例如 [m,n]=size(a) 如果函数的返回参数只有一个，中括号一般省略。即 [n]=rand(1,2); 可写�? n=rand(1,2); 大括��L��作用{ } 大括��P��用于cell型的数组的分配或引用�?br />比如 A(2,1) = {[1 2 3; 4 5 6]}, or A{2,2} = ('str') 分号�Q�的作用�Q?br /> cell�?数据�c�d�� 如果p��Z��个数�Q�那么h(1)=p,是没有问题的�?/p> 如果p��Z��个向量，那么h(1,:)=p是没有问题的�?/p> 如果p是一个矩�늚�话，上面的两�U�赋值方法都是会有错误的�?/p> 那么要如何处理呢�Q?/p> �q�时��q��C��cell数据�c�d��了。cell的每个单元都可以存储��M��数据�Q�比如传递函数等。当�Ӟ��存储矩阵更是没有问题的了。但是用cell数据�c�d��之前�Q�要先初始化�?/p> a=cell(n,m) 那么��把a初始化�ؓ一个n行m列的�I�cell�c�d��数据�?/p> 如何赋值呢�Q?/p> a{1,1}=rand(5) 那么a�?�?列的单元中存储的��是一个随机的5×5的方阵了�?/p> 那么要用�W�一个单元中的方阵中的某个值呢�Q?/p> 可以如下引用�Q�a{1,1}(2,3) ��可以了�Q�引用cell单元时要用{},再引用矩�늚�某个数据��p��?)了�?/p> cell单元中的每个单元都是独立的，可以分别存储不同大小的矩阉|��不同�c�d��的数据�?/p> luis 2012-10-11 08:28 发表评论 luis — Wed, 19 Sep 2012 01:47:00 GMT Python array 用法直接 result=[] for x in range(0,N): temp=beta(b,n) print temp if temp >= n: result.append("Yes") #直接append else: result.append("No") #直接append return result luis 2012-09-19 09:47 发表评论 Python�W�记 luis — Sun, 09 Sep 2012 06:47:00 GMT Tutorial �Q?/span> http://www.tutorialspoint.com/python/python_files_io.htm Python IO 输出 print str = raw_input("Enter your input: "); print "Received input is : ", str Python只有三种变量�c�d�� int, string, float? typeof(1.5) 貌似不支持隐式类型�{�?br />print str(2.5) �?br />print 2.5 �?br />print '2.5' �?br /> python 定义�Ҏ��?br />def MethodName(para,para2): #注意�q�里的冒�?br /> if python注释 #单行注释 """ 三个双引��h��多行注释 """ python 引用 include math Python的str str(var)�c�d��转换 len(var) var.upper() var.lower() var[2] �W�三个（注意下标�?开始）元素�Q�类��g��list var[:3] 前三个元素，实际上指的是0截止�?-1的元�?br />var[2:4]下标�?�?-1的所有元�?br /> Python的list exampe=[a,b,c,d,e,f]; len(exampe) 自带sort�Ҏ�� Python的dictionary key -value对应 value可以是一个list 注意�Ҏ��号[]里面只能使用key 区分 del dict['Name'] del dict['Name']; # remove entry with key 'Name' dict.clear(); # remove all entries in dict del dict ; # delete entire dictionary 自带的方法： 1 cmp(dict1, dict2) Compares elements of both dict. 2 len(dict) Gives the total length of the dictionary. This would be equal to the number of items in the dictionary. 3 str(dict) Produces a printable string representation of a dictionary 4 type(variable) Returns the type of the passed variable. If passed variable is dictionary then it would return a dictionary type. Python�cȝ��定义和类�Ҏ��的定�?br />定义�c�M��需要用def class,直接 class ClassName(object): 每个�c�都�?__init__(self,arg): �Ҏ��Q�注意是左右各两个下划线�Q��d��4根下划线 �c�L��法都需要包含self�q�个参数�Q�但是��用的时候不需要self,见下�?br /> 例如 class Adder(object): def __init__(self): self.baseNum = 2 def prnt_num(self): print self.baseNum def add_to_base(self, arg): # Your code here self.baseNum+=arg print self.baseNum objectVar = Adder() objectVar.prnt_num() # Your code here objectVar.add_to_base(3) Python中的�c�d��量不�?self.xxx来引用，但是成员变量可以 Class variables are special because they belong to the class; the objects created do not get their own copies of the class variable. Class variables are accessed using the class name and dot notation. ClassName.classVar Class variables are created outside of__init__ 例如�Q?br />class Widget(object): objID = 0 def __init__(self): Widget.objID += 1 # Your code here self.myID=Widget.objID 常犯错误�Q�indentation is very important python indentation error expected an indented block �q�有一个错误就�?�c�L��法，必须使用 self参数�Q�即使没有参敎ͼ��Q?/span> 另外一个常错的地方��是 __init__(self,arg) 一定是四根下划�U?/span> 不仅要记得留dent �q�要记得 �~�进 luis 2012-09-09 14:47 发表评论 luis — Tue, 28 Aug 2012 03:35:00 GMT 原地址�Q?a >http://blog.sina.com.cn/s/blog_5e16f1770100gzud.html 演示代码�Q?/p> \documentclass{article} \setlength\textwidth{245.0pt} \usepackage{CJK} \usepackage{indentfirst} \usepackage{amsmath} \begin{CJK*}{GBK}{song} \begin{document} �Ҏ��一�Q?/p> $$ f(x)=\left\{ \begin{aligned} x & = & \cos(t) \\ y & = & \sin(t) \\ z & = & \frac xy \end{aligned} \right. $$ �Ҏ��二： $$ F^{HLLC}=\left\{ \begin{array}{rcl} F_L & & {0 < S_L}\\ F^*_L & & {S_L \leq 0 < S_M}\\ F^*_R & & {S_M \leq 0 < S_R}\\ F_R & & {S_R \leq 0} \end{array} \right. $$ �Ҏ��? $$f(x)= \begin{cases} 0& \text{x=0}\\ 1& \text{x!=0} \end{cases}$$ \end{CJK*} \end{document} 可以用\text{} 注释文字 luis 2012-08-28 11:35 发表评论 Dijkstra(�q�杰斯特�?��法 luis — Fri, 15 Jun 2012 19:53:00 GMT 　Dijkstra(�q�杰斯特�?��法是典型的单源最短�\径算法，用于计算一个节点到其他所有节点的最短�\径。主要特�Ҏ��以�v始点��Z��心向外层层扩展，直到扩展到终点�ؓ止。Dijkstra��法是很有代表性的最短�\径算法，在很多专业课�E�中都作为基本内�Ҏ��详细的介�l�，如数据结构，图论�Q�运�{�学�{�等。Dijkstra一般的表述通常有两�U�方式，一�U�用�怹�和��时标��h��式，一�U�是用OPEN, CLOSE表的方式�Q�这里均采用�怹�和��时标��L��方式。注意该��法要求图中不存在负权边�?/span>问题描述　　在无向图 G=(V,E) 中，假设每条�?E[i] 的长度�ؓ w[i]�Q�找到由��点 V0 到其余各点的最短�\径。（单源最短�\�?/a>�Q?/span> 　�q�杰斯特�?/a>(Dijkstra)��法思想　　按�\径长度递增�ơ序产生最短�\径算法：　　把V分成两组�Q?/span> 　　�Q?�Q�S�Q�已求出最短�\径的��点的集�?/span> 　　�Q?�Q�V-S=T�Q�尚未确定最短�\径的��点集合　　��T中顶�Ҏ��最短�\径递增的次序加入到S中，　　保证�Q�（1�Q�从源点V0到S中各��点的最短�\径长度都不大�?/span> 　　从V0到T中�Q何顶点的最短�\径长�?/span> 　　�Q?�Q�每个顶点对应一个距��d�?/span> 　　S中顶点：从V0到此��点的最短�\径长�?/span> 　　T中顶点：从V0到此��点的只包括S中顶点作中间　　��点的最短�\径长�?/span> 　　依据�Q�可以证明V0到T中顶点Vk的最短�\径，或是从V0到Vk�?/span> 　　直接路径的权��|��或是从V0�l�S中顶点到Vk的�\径权��g��?/span> 　　�Q�反证法可证�Q?/span> 　　求最短�\径步�?/em> 　　��法步骤如下�Q?/span> 　　1. 初��时��o S={V0},T={其余��点}�Q�T中顶点对应的距离�?/span> 　　若存�?lt;V0,Vi>�Q�d(V0,Vi)�?lt;V0,Vi>弧上的权�?/span> 　　若不存在�Q�d(V0,Vi)�?#8733; 　　2. 从T中选取一个其距离��gؓ最��的��点W且不在S中，加入S 　　3. 对T中顶点的距离��D��行修改：若加�q�W作中间顶点，从V0到Vi�?/span> 　　距离值比不加W的�\径要短，则修�Ҏ��距离�?/span> 　　重复上述步骤2�?�Q�直到S中包含所有顶点，即S=T为止代码�Q?源地址�Q?span style="color: #008000; font-size: 13px; background-color: #eeeeee; ">www.cnblogs.com/newwy /********************************* * 最短�\�?--Dijkstra��法实现 * 　　　HDU�Q?544 * BLOG:www.cnblogs.com/newwy * AUTHOR:Wang Yong **********************************/ #include #define MAX 100 #define INF 1000000000 using namespace std; int dijkstra (int mat[][MAX],int n, int s,int f) { int dis[MAX]; int mark[MAX];//记录被选中的结�?nbsp; int i,j,k = 0; for(i = 0 ; i < n ; i++)//初始化所有结点，每个�l�点都没有被选中 mark[i] = 0; for(i = 0 ; i < n ; i++)//��每个结点到start�l�点weight记录为当前distance { dis[i] = mat[s][i]; //path[i] = s; } mark[s] = 1;//start�l�点被选中 //path[s] = 0; dis[s] = 0;//��start�l�点的的距离讄��? int min ;//讄��最短的距离�?nbsp; for(i = 1 ; i < n; i++) { min = INF; for(j = 0 ; j < n;j++) { if(mark[j] == 0 && dis[j] < min)//未被选中的结点中�Q�距��L��短的被选中 { min = dis[j] ; k = j; } } mark[k] = 1;//标记��选中 for(j = 0 ; j < n ; j++) { if( mark[j] == 0 && (dis[j] > (dis[k] + mat[k][j])))//修改剩余�l�点的最短距��?nbsp; { dis[j] = dis[k] + mat[k][j]; } } } return dis[f]; } int mat[MAX][MAX]; int main() { int n,m; while(scanf("%d %d",&n,&m)) { int a,b,dis; if(n == 0 || m == 0) break; int i,j; for(i = 0 ; i < n;i++) for(j = 0 ; j < n; j++) mat[i][j] = INF; for(i = 0 ; i < m ;i++) { scanf("%d %d %d",&a,&b,&dis); --a,--b; if(dis < mat[a][b] || dis < mat[b][a]) mat[a][b] = mat[b][a] = dis; } int ans = dijkstra(mat,n,0,n-1); printf("%d\n",ans); } } 可用优先队列优化其他解释�Q?br />http://blog.csdn.net/jiahui524/article/details/6636913 luis 2012-06-16 03:53 发表评论

1	cmp(dict1, dict2) Compares elements of both dict.
2	len(dict) Gives the total length of the dictionary. This would be equal to the number of items in the dictionary.
3	str(dict) Produces a printable string representation of a dictionary
4	type(variable) Returns the type of the passed variable. If passed variable is dictionary then it would return a dictionary type.

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原文地址: http://blog.csdn.net/gfaiswl/article/details/4749713

1.问题定义

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