青青草原综合久久大伊人导航_色综合久久天天综合_日日噜噜夜夜狠狠久久丁香五月_热久久这里只有精品

DraculaW

  C++博客 :: 首頁 :: 新隨筆 :: 聯系 :: 聚合  :: 管理 ::
  19 隨筆 :: 0 文章 :: 7 評論 :: 0 Trackbacks

2008年11月28日 #

     摘要: 我嘗試寫的一個智能指針
希望大家能幫我提一些意見 謝謝

上面是測試碼
下面是代碼  閱讀全文
posted @ 2008-11-28 17:39 DraculaW 閱讀(422) | 評論 (0)編輯 收藏

2007年11月22日 #

     摘要: 紅黑樹的插入算法的描述 不知道清楚么 希望大家指正  閱讀全文
posted @ 2007-11-22 20:47 DraculaW 閱讀(332) | 評論 (0)編輯 收藏

2007年11月21日 #

     摘要: 對內存分配算法的一點想法 希望大家指正  閱讀全文
posted @ 2007-11-21 22:38 DraculaW 閱讀(1994) | 評論 (1)編輯 收藏

2007年11月20日 #

     摘要: 大概的介紹了下deque的實現
為什么要用這種實現方式 誰能解釋一下  閱讀全文
posted @ 2007-11-20 22:30 DraculaW 閱讀(1192) | 評論 (0)編輯 收藏

2007年11月15日 #

template<typename T, typename U>
class Conversion
{
    typedef char Small;
    class Big{char dummy[2];};
    static Small Test(U)    {   }  
    static Big Test(...) { }  
    static T MakeT() { }
   
public:
    enum { exists = sizeof(Test(MakeT())) == sizeof(Small)};   
};

如果T可以轉換為U 于是就調用Test(U)這個函數 返回一個char;
如果不能 就調用使用(...)缺省參數的函數 返回一個數組

然后對返回值進行判斷....
posted @ 2007-11-15 20:45 DraculaW 閱讀(289) | 評論 (0)編輯 收藏

     摘要: STL與boost的type traits  閱讀全文
posted @ 2007-11-15 20:43 DraculaW 閱讀(954) | 評論 (0)編輯 收藏

# The readfpl accept a file's path while is fpl(foobar play list),
# and return a list which holds all the file'path
sub readfpl
{
    my @files;
    my @chunks;
    my $index = 0;

    open(INPUT, "< $_[0]")
        or die "can't open";

    @chunks = split(m{file://}, <INPUT>);

    foreach(@chunks)
    {   
        if($_ =~ m/.+\.(mp3|wma|m4a)/)
        {
            $files[$index] = $&;
            $index ++;
        }
    }

    print $files[0];
   
    return @files;
}

my @files = readfpl($ARGV[0]);
my $string;

foreach(@files){
     $string = $_;
        # the next while get name from path
     while( substr($string, 1) =~ m{\\.+\.(mp3|wna|m4a)}) {
         $string = $ARGV[0].$&;
         }
     rename $_, string;
}

將這段代碼存為movefpl.pl然后在命令行打入 movefpl.pl 播放列表的全路徑 要存歌曲的新路徑
就可以了呢 呵呵
posted @ 2007-11-15 20:41 DraculaW 閱讀(320) | 評論 (0)編輯 收藏

Assessing Infection

Background

According to the World Health Organization, infectious disease ranks as the leading cause of death in the world. In 1998 alone, over 17 million people died from infectious and parasitic diseases such as acute lower respiratory infections, tuberculosis, HIV/AIDS, and malaria. It is forecast that infectious disease will continue to kill millions of people, especially those living in developing countries.

The medical profession and scientific community of the world are fighting the infectious disease threat with new tools and technologies from a variety of fields. From this effort, a new field of research has emerged. Infectious Disease Epidemiology is the study of the variables that influence the growth and spread of infectious diseases. This relatively new field combines molecular biology, immunology, genetics, and the computational sciences. A focus of this field is the study of the factors that influence the growth of an infectious disease within a single organism, and the factors that influence the pattern of infection across an entire population.

Description

This assignment asks you to finish the implementation of a program that assesses the level of infection in a tissue sample. You are given data representing a rectangular tissue sample, overlaid with a grid. Certain portions of the tissue are infected; others are not. Your goal is to help assess the extent of the infection by writing a program that, given the coordinates of a colony of infection, can determine its size.

A typical use of the program follows. The user interacts with the program only through command-line arguments. The user supplies to the program a data filename and the coordinates of a cell in the grid. The coordinates are specified by row and then column, both starting at zero. The program calculates the extent of infection at that coordinate and outputs a two-dimensional representation of the tissue sample. Figure 1 depicts the execution of the program.

A screen shot from a sample solution
Figure 1 Output from a sample solution

For the purpose of this assessment, we consider a "colony" of infected tissue to be a set of adjacent and infected cells. In Figure 1, we can see three separate colonies. The smallest colony consists of two cells and is located in the lower left corner of the grid. Another colony consisting of three infected cells exists on the far right edge of the grid. The largest colony of eight cells resides primarily in the middle of the grid. This colony has a small arm into the upper left corner of the grid. Notice from this colony that cells residing in diagonals are considered "adjacent." The plus signs next to the cells in this largest colony indicate that they all belong to the colony that contains the user entered coordinate.


solution :

#ifndef GRID_H
#define GRID_H

#include <string>
#include <vector>

using namespace std;

/*
* IMPORTANT NOTE:
*
* For this assignment, you might need to add state to the
* class and/or augment existing methods, and/or create private helper
* methods, but you should not delare new public methods
*/

const bool INFECTED = true;
const bool NOT_INFECTED = false;

class grid;

class grid {

private:
    int rows;
    int cols;
    vector<bool> *area;
    vector<bool> *infect;
    int indexof (int row, int col) const;
    bool infected(int row, int col) const;

public:
    grid (string file);
    ~grid ();

    int count (int row, int col);

    friend ostream &operator<<(ostream &stream, const grid& ob);

};

#endif

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

#include <iostream>
#include <fstream>

using namespace std;

#include "grid.h"

// You do not need to alter function indexof.
int grid::indexof (int row, int col) const {
    return row*cols+col;
}

// You do not need to alter function infected.
bool grid::infected(int row, int col) const {
    return (area->operator[](indexof(row, col)) == INFECTED);
}

// You may need to alter the constructor
grid::grid (string file) {

    ifstream grid_file;

    grid_file.open (file.c_str());

    grid_file >> rows;
    grid_file >> cols;

    area = new vector<bool>(rows*cols, NOT_INFECTED);
    infect = new vector<bool>(rows*cols, NOT_INFECTED);
   
    while (true) {

        int blob_row;
        int blob_col;

        grid_file >> blob_row;
        grid_file >> blob_col;

        if (grid_file.eof()) {
            break;
        }

        area->operator[](indexof(blob_row,blob_col)) = INFECTED;
    }

    grid_file.close();
}

// You may need to alter the destructor
grid::~grid () {
    delete area;
    delete infect;
}

// You will need to alter this function to display the
// plus signs (+) next to the cells that belong to
// a counted colony.
ostream &operator<<(ostream &stream, const grid& ob) {

    for (int row=0; row < ob.rows; row++) {
   
        for (int col=0; col < ob.cols; col++) {

            stream << ob.area->operator[](ob.indexof(row, col));
            if( ob.infect->operator[] ( ob.indexof(row, col) ) )
                stream << "+ ";
            else
                stream << "   ";
        }

        stream << endl;
    }

    stream << endl;
    return stream;
}

// Replace the return statement in this function with your
// recursive implementation of this method */
int grid::count (int row, int col) {

    if( row < 0 || col < 0 || row == rows || col == cols)
        return 0;

    if( area->operator[](indexof(row,col) ) == NOT_INFECTED )
        return 0;

    if(infect->operator[](indexof(row,col)) == INFECTED)
        return 0;

    infect->operator[](indexof(row,col)) = INFECTED;

    // Recursive test the 8 point near the point
    // which area is INEFCTED and infect is NOT_INFECTED

    return    count( row - 1, col - 1 ) + count ( row - 1, col )
        + count( row - 1, col + 1 ) + count( row, col - 1 )
        + count( row, col ) + 1 + count( row, col + 1 )
        + count( row + 1, col - 1 ) + count( row + 1, col )
        + count( row + 1, col + 1 );
}
posted @ 2007-11-15 20:40 DraculaW 閱讀(508) | 評論 (0)編輯 收藏

     摘要: 一道簡單的作業貼 呵呵   閱讀全文
posted @ 2007-11-15 20:39 DraculaW 閱讀(472) | 評論 (0)編輯 收藏

其實這個題目也很簡單 有很多種做法...

就是給一個array你 然后你找出 i,j使從第i個加到第j個最大就好了啊

最簡單的算法就是兩個for 算下來不到n^2的時間復雜度 可是還有更快的算法哦

首先 可以使用分治算法 這樣的算法大概時間復雜度是 n*lg n, 但是這樣還不是最好的

最好的其實是把前一個狀態儲存下來然后進行比較 這個算法時間復雜度只有n哦 很快的呢

先不要看 給個 int a[10] = { 31, -41, 59, 26, -53, 58, 97, -93, -23, 84 }

求它的最大子串有多大哦

inline int
max( int a, int b)
{
    return a > b ? a : b;
}

/*****************************************************************************
* This Function count a array find the largest string count max              *
* Function : CountMax                                                        *
* int    *a : the array of int                                                *
* int     n : the range of array                                              *
* return    : the sum of max this function find                               *
*****************************************************************************/
int
CountMax ( int *a, int n )
{
    int sum = 0, tmp = 0;
    for( int i = 0; i < n; i++ )
    {
        tmp = max( 0, tmp + a[i] );
        sum = max( sum, tmp );
    }

    return sum;
}
/* -----   end of function CountMax   ----- */
posted @ 2007-11-15 20:37 DraculaW 閱讀(160) | 評論 (0)編輯 收藏

僅列出標題  下一頁
青青草原综合久久大伊人导航_色综合久久天天综合_日日噜噜夜夜狠狠久久丁香五月_热久久这里只有精品
  • <ins id="pjuwb"></ins>
    <blockquote id="pjuwb"><pre id="pjuwb"></pre></blockquote>
    <noscript id="pjuwb"></noscript>
          <sup id="pjuwb"><pre id="pjuwb"></pre></sup>
            <dd id="pjuwb"></dd>
            <abbr id="pjuwb"></abbr>
            午夜精品福利在线| 久久日韩精品| 国产精品久久久亚洲一区| 亚洲网友自拍| 亚洲天堂av电影| 国产精品无码专区在线观看| 亚洲视频免费在线| 亚洲一区精品在线| 国内成+人亚洲| 欧美高清免费| 欧美日韩一区二区在线| 亚洲影院色在线观看免费| 午夜激情综合网| 在线视频成人| 中日韩美女免费视频网址在线观看| 国产精品国产三级欧美二区| 欧美制服丝袜第一页| 久久久亚洲精品一区二区三区| 在线色欧美三级视频| av成人福利| 精品96久久久久久中文字幕无| 亚洲国产你懂的| 国产精品午夜春色av| 久久综合图片| 国产精品久久久久久久久果冻传媒 | 免费成人在线观看视频| 欧美va天堂| 欧美一区成人| 欧美精品一区在线| 久久精品国产v日韩v亚洲| 欧美~级网站不卡| 香蕉久久一区二区不卡无毒影院 | 欧美在线免费播放| 99re热这里只有精品免费视频| 亚洲在线日韩| 99视频在线精品国自产拍免费观看| 一区二区三区久久| 最新高清无码专区| 欧美在线观看视频一区二区| 亚洲视频1区| 久久这里有精品视频| 午夜在线一区| 欧美四级剧情无删版影片| 能在线观看的日韩av| 国产人成一区二区三区影院| 亚洲国产综合在线看不卡| 韩国女主播一区| 亚洲一区二区三区在线| 一区二区三区黄色| 欧美 日韩 国产在线| 久久久亚洲人| 国产视频久久久久久久| 亚洲视频香蕉人妖| 一区二区三区.www| 欧美激情一区在线| 欧美激情一区三区| 亚洲成色www久久网站| 亚洲欧美综合另类中字| 先锋a资源在线看亚洲| 欧美性猛交视频| 日韩系列欧美系列| 一区二区免费在线视频| 欧美精品在线视频| 亚洲激情黄色| 99精品视频免费观看| 欧美国产欧美综合 | 久久福利一区| 国产一区亚洲| 欧美综合国产精品久久丁香| 久久激情中文| 激情五月婷婷综合| 老色鬼久久亚洲一区二区| 欧美成人免费小视频| 亚洲电影网站| 欧美国产精品| 中文在线一区| 久久久精彩视频| 樱桃成人精品视频在线播放| 久久久水蜜桃| 亚洲人体偷拍| 亚洲永久免费视频| 国产午夜亚洲精品理论片色戒| 欧美一区二区三区日韩| 久久综合五月天婷婷伊人| 亚洲国产成人在线播放| 欧美激情亚洲激情| 亚洲视频图片小说| 久久久久久成人| 亚洲黄色片网站| 欧美日韩直播| 欧美一区二区三区四区视频| 免费欧美日韩| 亚洲综合日韩在线| 国产综合第一页| 欧美91大片| 亚洲一区二三| 免费成人黄色| 亚洲视频1区| 激情久久久久久| 欧美日韩另类一区| 欧美一区二区日韩| 亚洲日本激情| 久久精品视频免费播放| 亚洲人成在线影院| 国产嫩草一区二区三区在线观看 | 欧美精品久久99久久在免费线| 亚洲午夜一区| 欧美激情偷拍| 久久aⅴ国产紧身牛仔裤| 亚洲乱码视频| 黄色成人91| 国产精品嫩草影院一区二区 | 最近看过的日韩成人| 欧美一区亚洲二区| 亚洲最新中文字幕| 激情久久久久久| 国产精品嫩草影院av蜜臀| 欧美成人高清视频| 欧美中文字幕在线视频| 亚洲国产日韩欧美在线图片| 久久激五月天综合精品| 国产精品99久久久久久人| 亚洲国产精品久久久久秋霞蜜臀 | 国产精品一二三| 欧美日韩一区二区三区| 久久综合久色欧美综合狠狠| 亚洲午夜av在线| 亚洲欧洲一区二区三区久久| 久久久久久久久久久久久女国产乱 | 老司机精品久久| 欧美一区二区在线看| 夜夜精品视频一区二区| 亚洲电影有码| 在线观看一区视频| 好吊妞这里只有精品| 国产欧美精品久久| 国产精品视频一区二区三区| 欧美区在线观看| 欧美精选一区| 欧美日本乱大交xxxxx| 免费欧美高清视频| 开元免费观看欧美电视剧网站| 新67194成人永久网站| 亚洲一区视频在线观看视频| 一区二区三区精品视频| 一区二区久久| 亚洲一区国产精品| 午夜亚洲福利在线老司机| 亚洲欧美成人精品| 亚洲欧美一区二区视频| 性xx色xx综合久久久xx| 香蕉av777xxx色综合一区| 欧美一区二区三区日韩视频| 欧美一区二区在线看| 久久激情五月丁香伊人| 久久三级视频| 欧美成ee人免费视频| 欧美日韩色一区| 国产精品日韩在线观看| 国产精品亚发布| 黄色成人在线免费| 亚洲国产另类精品专区| 亚洲精品视频免费| 亚洲综合精品四区| 久久免费视频在线| 欧美成人69av| 亚洲美女精品久久| 性色av一区二区三区红粉影视| 欧美亚洲一区在线| 欧美国产另类| 国产精品一区二区三区四区五区 | 国产精品国产三级国产普通话99| 国产欧美在线播放| 亚洲啪啪91| 亚洲欧美国产77777| 久久大逼视频| 亚洲国产成人av| 亚洲欧美日韩在线播放| 另类成人小视频在线| 欧美午夜精品理论片a级大开眼界| 国产毛片精品国产一区二区三区| 激情视频一区二区| 亚洲图片在线| 欧美电影免费观看高清| 亚洲视频综合在线| 欧美a级片网站| 国产偷国产偷精品高清尤物| 91久久精品一区二区别| 久久本道综合色狠狠五月| 亚洲福利免费| 欧美专区亚洲专区| 国产精品99免费看 | 午夜精品一区二区三区在线视| 久久久久网站| 欧美视频网站| 亚洲伦理自拍| 麻豆成人综合网| 欧美一区二区在线免费观看| 欧美三级网址| 亚洲精品一区二区三区不|