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HDU 3468 HDOJ 3468 A Simple Problem with Integers ACM 3468 IN HDU

MiYu — Tue, 16 Nov 2010 13:28:00 GMT

摘要: MiYu原创, 转帖��h��?: 转蝲�?nbsp;______________白白の屋代码/*Mail to : miyubai@gamil.comMy Blog :&nb... 阅读全文

MiYu 2010-11-16 21:28 发表评论

PKU 2528 POJ 2528 Mayor's posters ( �U�段�?��L��?) ACM 2528 IN PKU

MiYu — Fri, 15 Oct 2010 03:12:00 GMT

摘要: MiYu原创, 转帖��h��?: 转蝲�?nbsp;______________白白の屋题目地址 : http://poj.org/problem?id=2528题目描述:Mayor's postersTime Limit: 1000MSMemory Limit: 65536KTotal Submissions: 1572... 阅读全文

MiYu 2010-10-15 11:12 发表评论

PKU 2352 POJ 2352 Stars ( �U�段树版 ) ACM 2352 IN PKU

MiYu — Wed, 13 Oct 2010 14:29:00 GMT

MiYu原创, 转帖��h��?: 转蝲�?nbsp;______________白白の屋

题目地址 :

http://poj.org/problem?id=2352

题目描述:

Stars
Time Limit: 1000MSMemory Limit: 65536K
Total Submissions: 14814Accepted: 6404
Description
Astronomers often examine star maps where stars are represented by points on a plane and each star has Cartesian coordinates. Let the level of a star be an amount of the stars that are not higher and not to the right of the given star. Astronomers want to know the distribution of the levels of the stars. 

For example, look at the map shown on the figure above. Level of the star number 5 is equal to 3 (it's formed by three stars with a numbers 1, 2 and 4). And the levels of the stars numbered by 2 and 4 are 1. At this map there are only one star of the level 0, two stars of the level 1, one star of the level 2, and one star of the level 3. 

You are to write a program that will count the amounts of the stars of each level on a given map.
Input
The first line of the input file contains a number of stars N (1<=N<=15000). The following N lines describe coordinates of stars (two integers X and Y per line separated by a space, 0<=X,Y<=32000). There can be only one star at one point of the plane. Stars are listed in ascending order of Y coordinate. Stars with equal Y coordinates are listed in ascending order of X coordinate. 
Output
The output should contain N lines, one number per line. The first line contains amount of stars of the level 0, the second does amount of stars of the level 1 and so on, the last line contains amount of stars of the level N-1.
Sample Input
5
1 1
5 1
7 1
3 3
5 5Sample Output
1
2
1
1
0 与树状数�l�的解法一�?  查询更新���可以了. 	树状数组 解法传送门 : http://www.cnblogs.com/MiYu/archive/2010/08/26/1808963.html 代码如下 : /*
Mail to   : miyubai@gamil.com
Link      : http://www.cnblogs.com/MiYu  || http://www.shnenglu.com/MiYu
Author By : MiYu
Test      : 1
Complier  : g++ mingw32-3.4.2
Program   : POJ_2352
Doc Name  : Stars
*/
//#pragma warning( disable:4789 )
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
struct P{
       int x, y;
}p[15010];
int seg[100000];
int NN = 32768, res[15010];
int get ( int &n ) {
    int i = 1;
    while ( i < n ) i << 1;
    return i;    
}
void modify ( int &pos ) {
     pos += NN;
     seg[pos] ++;
     while ( pos > 1 ) {
           if ( ~pos & 1 ) seg[pos>>1] ++;
           pos >>= 1;
     }
}
int query ( int pos ) {
     pos += NN;  
     int sum = seg[pos];
     while ( pos > 1 ) {
            if ( pos & 1 ) {
                 sum += seg[pos>>1];  
            }      
            pos >>= 1;
     }   
     return sum;
}
int main ()
{
    int N;
    while ( scanf ( "%d", &N ) == 1 ) {
           int pos;
           memset ( res, 0, sizeof ( res ) );
           memset ( seg, 0, sizeof ( seg ) );
           for ( int i = 0; i < N; ++ i ) {
                  scanf ( "%d%*d", &pos );
                  res[ query ( pos ) ] ++;
                  modify ( pos );   
           }   
           for ( int i = 0; i < N; ++ i ) {
                printf ( "%d\n", res[i] );     
           } 
    }
    return 0;
}

MiYu 2010-10-13 22:29 发表评论

HDU 3016 HDOJ 3016 Man Down ACM 3016 IN HDU

MiYu — Wed, 13 Oct 2010 13:26:00 GMT

MiYu原创, 转帖��h��?: 转蝲�?nbsp;______________白白の屋

题目地址 :

http://acm.hdu.edu.cn/showproblem.php?pid=3016

题目描述:

Man DownTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 618    Accepted Submission(s): 197


Problem Description
The Game “Man Down 100 floors” is an famous and interesting game.You can enjoy the game from 
http://hi.baidu.com/abcdxyzk/blog/item/16398781b4f2a5d1bd3e1eed.html


We take a simplified version of this game. We have only two kinds of planks. One kind of the planks contains food and the other one contains nails. And if the man falls on the plank which contains food his energy will increase but if he falls on the plank which contains nails his energy will decrease. The man can only fall down vertically .We assume that the energy he can increase is unlimited and no borders exist on the left and the right.

First the man has total energy 100 and stands on the topmost plank of all. Then he can choose to go left or right to fall down. If he falls down from the position (Xi,Yi),he will fall onto the nearest plank which satisfies (xl <= xi <= xr)(xl is the leftmost position of the plank and xr is the rightmost).If no planks satisfies that, the man will fall onto the floor and he finishes his mission. But if the man’s energy is below or equal to 0 , he will die and the game is Over.

Now give you the height and position of all planks. And ask you whether the man can falls onto the floor successfully. If he can, try to calculate the maximum energy he can own when he is on the floor.(Assuming that the floor is infinite and its height is 0,and all the planks are located at different height).
 

Input
There are multiple test cases.

For each test case, The first line contains one integer N (2 <= N <= 100,000) representing the number of planks.

Then following N lines representing N planks, each line contain 4 integers (h,xl,xr,value)(h > 0, 0 < xl < xr < 100,000, -1000 <= value <= 1000), h represents the plank’s height, xl is the leftmost position of the plank and xr is the rightmost position. Value represents the energy the man will increase by( if value > 0) or decrease by( if value < 0) when he falls onto this plank.
 

Output
If the man can falls onto the floor successfully just output the maximum energy he can own when he is on the floor. But if the man can not fall down onto the floor anyway ,just output “-1”(not including the quote)
 

Sample Input

4
10 5 10 10
5 3 6 -100
4 7 11 20
2 2 1000 10

 

Sample Output

140

 

 /*
  题目描述:  
          不同高度处有不同的水�q�x���Q�蟩到该板会有血量变化v�Q?/p>
          问当一个�h从最高板开始，可以向左或者向叻I��
          竖直跛_��下面的板�Q�求下落到地面的最大血量，或�?1�?/p>
          �U�段�?dp  
          需要用�U�段树查询得到每个板的两个端点下落后会到哪个板；
          然后�Ҏ(gu��)���q�个从最高的开始dp���可以了
          dp[i] = max ( dp[i], dp[i^].v )  // dp[i^] 代表能走�?i 的线�D?nbsp;
/* 

/*
Mail to   : miyubai@gamil.com
Link      : http://www.cnblogs.com/MiYu  || http://www.shnenglu.com/MiYu
Author By : MiYu
Test      : 1
Complier  : g++ mingw32-3.4.2
Program   : HDU_3016
Doc Name  : Man Down
*/
//#pragma warning( disable:4789 )
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
struct seg_tree {
    int id, left, right;
    int mid () { return ( left + right )>>1; }  
}seg[333333];
inline void creat ( int x, int y, int rt = 1 ) {
     seg[rt].left = x;
     seg[rt].right = y;
     //0 代表地面 其他的自然数代表各层的木板编�? -1 代表有多条线�D�覆�?nbsp;
     seg[rt].id = 0;                  
     if ( x == y ) return ;
     int mid = seg[rt].mid();
     creat ( x, mid, rt << 1 );
     creat ( mid + 1, y, rt << 1 | 1 );     
}
inline void modify ( int x, int y, int id, int rt = 1 ) {
     //扑ֈ�了线�D? 直接修改ID 覆盖�?nbsp;
     if ( seg[rt].left == x && seg[rt].right == y ) {
         seg[rt].id = id;
         return;   
     }
     int LL = rt << 1, RR = rt << 1 | 1, mid = seg[rt].mid();
     // 前面没有return�? 那么�q�条�U�段肯定是被覆盖�? ���它的标��C��传后标记�?1 
     if ( seg[rt].id != -1 ) {      
         seg[LL].id = seg[RR].id = seg[rt].id;          
         seg[rt].id = -1;
     }      
     if ( y <= mid ) modify ( x, y, id, LL ); //分段修改 
     else if ( x > mid ) modify ( x, y, id, RR );
     else {
          modify ( x, mid, id, LL );
          modify ( mid + 1, y, id, RR );     
     }
}
inline int query ( int pos, int rt = 1 ) {   // 查询 Pos 所在线�D늚� id  
    if ( seg[rt].id != -1 ) return seg[rt].id; //�U�段被覆�?直接�q�回 ID 
    int LL = rt << 1, RR = rt << 1 | 1, mid = seg[rt].mid();
    if ( pos <= mid ) return query ( pos, LL );             //分段查询 
    else return query ( pos, RR );    
}
inline bool scan_d(int &num)  //整数输入
{
        char in;bool IsN=false;
        in=getchar();
        if(in==EOF) return false;
        while(in!='-'&&(in<'0'||in>'9')) in=getchar();
        if(in=='-'){ IsN=true;num=0;}
        else num=in-'0';
        while(in=getchar(),in>='0'&&in<='9'){
                num*=10,num+=in-'0';
        }
        if(IsN) num=-num;
        return true;
}
struct Plank {
       int x,y,h,v,left,right; 
       //按高排序       
       friend bool operator < ( const Plank &a, const Plank &b ) {
              return a.h < b.h;
       }
}pk[100010];
int dp[100010];
int main ()
{
    int N, M;
    creat ( 1, 100000 );
    while ( scan_d ( N ) ) {
           M = -1;
           for ( int i = 1; i <= N; ++ i ) {
                scan_d ( pk[i].h );scan_d ( pk[i].x );scan_d ( pk[i].y );scan_d ( pk[i].v );
                if ( pk[i].y > M ) M = pk[i].y;       // 记录 区间最大�? 加速用�?nbsp;
           }
           modify  ( 1, M, 0 );
           sort ( pk + 1, pk + N + 1 );               // 按高排序 
           memset ( dp, 0, sizeof ( dp ) );
           dp[N] = 100 + pk[N].v;
           // 自底向上 更新 �U�段, 记录 每条�U�段 左右端点能到辄��� �U�段 ID 
           for ( int i = 1; i <= N; ++ i ) {          
                int x = pk[i].left = query ( pk[i].x );
                int y = pk[i].right = query ( pk[i].y );
                modify ( pk[i].x, pk[i].y, i );
           }
           int res = -1;
           //自顶向下 DP    dp[i] = max ( dp[i], dp[i^].v )  
           // dp[i^] 代表能走�?i 的线�D?nbsp;
           for ( int i = N; i >= 1; -- i ) {   
               if ( dp[ pk[i].left ] < dp[i] + pk[ pk[i].left ].v )
                    dp[ pk[i].left ] = dp[i] + pk[ pk[i].left ].v;
               if ( dp[ pk[i].right ] < dp[i] + pk[ pk[i].right ].v )
                    dp[ pk[i].right ] = dp[i] + pk[ pk[i].right ].v; 
           } 
           printf ( "%d\n",dp[0] > 0 ? dp[0] : -1 );
    }
    return 0;
}

 

MiYu 2010-10-13 21:26 发表评论

HDU 2871 HDOJ 2871 Memory Control ACM 2871 IN HDU

MiYu — Thu, 07 Oct 2010 13:49:00 GMT

摘要: MiYu原创, 转帖��h��?: 转蝲�?nbsp;______________白白の屋题目地址: http://acm.hdu.edu.cn/showproblem.php?pid=2871题目描述:Memory ControlTime Limit: 2000/1000 MS (Java/Others) ... 阅读全文

MiYu 2010-10-07 21:49 发表评论

PKU 3667 HDOJ 3667 Hotel ACM 3667 IN HDU

MiYu — Thu, 07 Oct 2010 09:36:00 GMT

摘要: MiYu原创, 转帖��h��?: 转蝲�?nbsp;______________白白の屋题目地址 : http://poj.org/problem?id=3667题目描述:HotelTime Limit: 3000MSMemory Limit: 65536KTotal Submissions: 2993Acce... 阅读全文

MiYu 2010-10-07 17:36 发表评论

HDU 1540 HDOJ 1540 Tunnel Warfare ACM 1540 IN HDU

MiYu — Wed, 06 Oct 2010 03:06:00 GMT

摘要: MiYu原创, 转帖��h��?: 转蝲�?nbsp;______________白白の屋题目地址: http://acm.hdu.edu.cn/showproblem.php?pid=1540题目描述:Tunnel WarfareTime Limit: 4000/2000 MS (Java/Others) ... 阅读全文

MiYu 2010-10-06 11:06 发表评论

�U�段树的一�U�简化实现[转] by �t�雪赤兔

MiYu — Mon, 04 Oct 2010 08:11:00 GMT

转蝲�?nbsp;http://www.cnitblog.com/cockerel/archive/2006/09/13/16806.html

�怿�对算法设计或者数据结构有一定了解的人对�U�段树都不会太陌生。它是能够在log(MaxLen)旉��内完成线�D늚��d��、删除、查询等操作。但一般的实现都有点复杂（我自写的是要递归的，比较多行�Q�。而线�D�|��应用中有一�U�是专门针对点的。（�Ҏ(gu��)��Q�）它的实现却非常简单�?br>　　�q�种数据�l�构有什么用�Q�我们先来考虑一下下面的需求（全部要求在LogN旉��内完成）�Q�如何知道一个点在一个点集里的大��?#8220;排名”�Q�很��单，开一个点数组�Q�排个序�Q�再二分查找��p��了；如何在一个点集内动态增删点�Q�也很简单，弄个�q��树就行了�Q�本来��^衡树比线�D�|��复杂得多�Q�但自从世界上有了STL set�q�么个好东东�Q�就……^_^�Q�那如果我既要动态增删点�Q�也要随时查询到一个点的排名呢�Q�那对不��P��可能��p��出动到我们的“�Ҏ(gu��)��”了�?br>　　其实现原理很��单：每当增加�Q�或删除�Q�一个大��ؓX的点�Ӟ��在树上��d��Q�或删除�Q�一�?X,MaxLen)的线�D�（不含端点�Q�，当要查询一个点的排名时�Q�只要看看其上有多少条线�D�就可以了。针对这一需求，�q�里有个非常��单的实现�Q�见以下代码�Q�十多行�Q�够短了吧？�Q�其中clear()用于清空炚w��Q�add()用于��d��一个点�Q�cntLs()�q�回��于n的点的个敎ͼ�也就是n的升序排名，�c�M��地cntGt是降序排名�?br>　　�q�个�Ҏ(gu��)��有什么用呢？其中一个应用时在O(NlogN)旉��内求��Z��个排列的逆序敎ͼ�http://acm.zju.edu.cn/show_problem.php?pid=1484�Q�你有更好的��法吗？�Ƣ迎交流�Q�方法是每读��C��个数x�Q�就让逆序�?=cntGt(x);然后再add(x)�?br>　　�q�个实现�q�可以进行一些扩展。比如删除del(int n)�Q�只要把add(int n)中的++size换成--size�Q�把a[i/2]++�Ҏ(gu��)��a[i/2]--卛_��。另外还可以通过二分查找功能在O(logN)旉��内查到排名第n的点的大��。应该也可以三四行内搞定�?/p>

template < int  N > // 表示可用区间为[0,N)�Q�其中N必须�?的幂�?
class  PointTree {
     int  a[ 2 * N];
     int  size;
     void  clear() { memset( this , 0 , sizeof ( * this ));}
     void  add( int  n) {
         int  i = N + n;  ++ size;
         for ( ++ a[i]; i > 1 ; i /= 2 )
             if ( ~ i & 1 ) a[i / 2 ] ++ ;
    }
     int  cntLs( int  n) { // �l�计��于
          int  i = N + n,c = 0 ;  // 若统计小于等于则c=a[i];
          for (; i > 1 ; i /= 2 )
             if (i & 1 ) c += a[i / 2 ];
          return  c;
    }
      int  cntGt( int  n) {  return  size - a[N + n] - cntLs(n); }
} ;

　　嗯~~~��Z��?/span>http://acm.zju.edu.cn/show_problem.php?pid=2425�q�一题，�q�是把上�q�C��个扩展给实现了啦�Q�果然不难：

�Q�接上）
    void del(int n){
        if(!a[n+=N])return;
        --size;
        for(--a[n]; n>1; n/=2)
            if(~n&1)--a[n/2];
    }
    /*  解决�Q�求炚w��中第i��的�?�?数�v)
     *    注意�Q�如果i>=size �q�回N-1
     */
    int operator[](int n){
        int i=1;
        while(i<N){
            if(n<a[i]) i*=2;
            else n-=a[i], i=i*2+1;
        }
        return i-N;
    }
};
//附一个测试程�?/span>
#include<iostream.h>
T<8192> t;
int main(){
    char c; int n;
    while(cin>>c){
           if(c=='c') t.clear();
           else{
               cin>>n;
               if(c=='a') t.add(n);
               if(c=='d') t.del(n);
               if(c=='q') cout<<t[n]<<endl;
           }
    }

    return 0;
}

求点集里n的个��C��!

int cntEQ(int n){
return a[N+n];
}

P.S.�Q?/span>
　　在写完这��文章一�D�|��间后�Q�我认识了另一�U�功能上比较�c�M��的数据结构，叫做“树状数组”。它们有不少�怼�之处�Q?/span>

针对炚w��的处理（��d��、删除、查找）�Q?/li>
�怼�的时�I�复杂度�Q�logN旉��Q?N�I�间�Q�；
�怼�的编�E�复杂度�Q�都比线�D�|��短得多）�Q?/li>

因此�Q�所有可以用树状数组解决的问题都可以用这�?#8220;�Ҏ(gu��)��”来解冻I��另外它还有以下好处：

更直观的转移�Q�个人感受，不一定要同意�Q�；
同时支持自下而上和自上而下两种方向的查扑֒�更新�Q�而后者树状数�l�不支持�Q�所以树状数�l�不提供某些功能�Q�比如说O(logN)求点集中�W�k��数�?/li>

posted on 2006-09-13 19:54 �t�雪赤兔

MiYu 2010-10-04 16:11 发表评论

HDU 2795 HDOJ 2795 Billboard ACM 2795 IN HDU

MiYu — Sun, 26 Sep 2010 14:15:00 GMT

MiYu原创, 转帖��h��?: 转蝲�?nbsp;______________白白の屋

题目地址:

http://acm.hdu.edu.cn/showproblem.php?pid=2795

题目描述:

BillboardTime Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 733    Accepted Submission(s): 340


Problem Description
At the entrance to the university, there is a huge rectangular billboard of size h*w (h is its height and w is its width). The board is the place where all possible announcements are posted: nearest programming competitions, changes in the dining room menu, and other important information.

On September 1, the billboard was empty. One by one, the announcements started being put on the billboard.

Each announcement is a stripe of paper of unit height. More specifically, the i-th announcement is a rectangle of size 1 * wi.

When someone puts a new announcement on the billboard, she would always choose the topmost possible position for the announcement. Among all possible topmost positions she would always choose the leftmost one.

If there is no valid location for a new announcement, it is not put on the billboard (that's why some programming contests have no participants from this university).

Given the sizes of the billboard and the announcements, your task is to find the numbers of rows in which the announcements are placed.
 

Input
There are multiple cases (no more than 40 cases).

The first line of the input file contains three integer numbers, h, w, and n (1 <= h,w <= 10^9; 1 <= n <= 200,000) - the dimensions of the billboard and the number of announcements.

Each of the next n lines contains an integer number wi (1 <= wi <= 10^9) - the width of i-th announcement.
 

Output
For each announcement (in the order they are given in the input file) output one number - the number of the row in which this announcement is placed. Rows are numbered from 1 to h, starting with the top row. If an announcement can't be put on the billboard, output "-1" for this announcement.
 

Sample Input
3 5 5
2
4
3
3
3
 

Sample Output
1
2
1
3
-1
 

 
一开始没��x��白怎么�?,  仔细想了�?   再次 读题�?发现 , n <= 200000;  也就是说  最�?也就 200000 条广�?, 你就���每行脓(chu��ng)一�?,
最多也���p��(chu��ng)�?200000 �?  所�? 不要�? h <= 10^9  �ơ方吓到�? ,认�ؓ �U�段树开不了那么大的数组  . 只要开 200000 ���可以了 .
 
其他�?没什�?好说�?, 知道�q�个 ���q���?�?�?............���近 7000MS .=水过.................   g++提交 �q�华丽的 送出�?一��?TLE....	
C++  水过�?............
 
//    一直没明白 ��Z��么我�?代码速度 那么 �?  查询�?更新 的时间是 2000MS 左右  , 我的 �?查询 ���更��C��,
竟然�?���近 7000MS ? 非常 郁闷 ,  不信邪的 �l�箋 ����?代码, �?瞪了 1哥小时后 , 忽然惛_�� : �?cout �Ҏ(gu��)��
printf 会怎样?   �l�果 :  1640 MS  AC ..............鬼知道他�?数据量有多大..... cout �?printf
竟然 差了 5000 MS 的时�?...........无语
代码如下 :
 
 /*
Coded By  : MiYu
Link      : http://www.cnblogs.com/MiYu  || http://www.shnenglu.com/MiYu
Author By : MiYu
Test      : 1
Program   : 2795
*/
//#pragma warning( disable:4789 )
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

struct ADV {
       int left, right, val;
       int mid () { return ( left + right ) >> 1; }     
}adv[800000];
int N, W, H, w;
void creat ( int l, int r, int rt = 1 ){
     adv[rt].left = l;
     adv[rt].right = r;
     adv[rt].val = W;
     if ( l == r )
        return ;
     int mid = adv[rt].mid();
     creat ( l, mid, rt << 1 );
     creat ( mid + 1, r, ( rt << 1 ) + 1 );
}
void add ( int rt = 1, int wei = w ){
    if ( wei <= adv[rt].val ){
         if ( adv[rt].left == adv[rt].right ){
              adv[rt].val -= wei;
              //cout << adv[rt].left << endl;  //杯具�?地方
	printf ( "%d\n", adv[rt].left ); 
              return ;     
         } else if ( adv[rt<<1].val >= wei ) {
                add ( rt << 1 );
                adv[rt].val = max ( adv[rt<<1].val, adv[(rt<<1)+1].val );     
         } else {
                add ( ( rt << 1 ) + 1 );
                adv[rt].val = max ( adv[rt<<1].val, adv[(rt<<1)+1].val );        
         }   
    } else {
           //cout << -1 << endl;      //杯具的地�?/span>
	puts ( "-1" );   
    }   
}
inline bool scan_ud(int &num)
{
    char in;
    in=getchar();
    if(in==EOF) return false;
    while(in<'0'||in>'9') in=getchar();
    num=in-'0';
    while(in=getchar(),in>='0'&&in<='9'){
        num*=10,num+=in-'0';
    }
    return true;
}
int main ()
{
    while ( scan_ud (H)&&scan_ud (W)&&scan_ud (N) ) {
           if ( H > 200000 )
               H = 200010;
           creat ( 1, H );
           for ( int i = 1; i <= N; ++ i ) {
                scan_ud (w);
                add ( );     
           }      
    }
    return 0;
}
 
 
 
 
�?附上 ��d��  ���牛 代码 :
 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define FF(i,a)				for( int i = 0 ; i < a ; i ++ )
#define FOR(i,a,b)			for( int i = a ; i < b ; i ++ )
#define LL(a)				a<<1		
#define RR(a)				a<<1|1		
template inline void checkmin(T &a,T b)	{if(a < 0 || a > b)a = b;}
template inline void checkmax(T &a,T b)	{if(a < b)	a = b;}
using namespace std;
struct Node {
	int val;
	int idx;
	friend bool operator < (Node a , Node b) {
		if(a.val == b.val) {
			return a.idx > b.idx;
		}
		return a.val < b.val;
	}
}error;
 
struct Seg_Tree{
	int left,right;
	Node node;
	int mid() {
		return (left + right)>>1;
	}
}tt[800000];
 
int n , h , m;
 
void build(int l,int r,int idx) {
	tt[idx].left = l;
	tt[idx].right = r;
	tt[idx].node.idx = l;
	tt[idx].node.val = h;
	if(l == r) return ;
	int mid = tt[idx].mid();
	build(l,mid,LL(idx));
	build(mid+1,r,RR(idx));
}
 
void update(int l,int r,int val,int idx) {
	if(l <= tt[idx].left && r >= tt[idx].right) {
		tt[idx].node.val += val;
		return ;
	}
	int mid = tt[idx].mid();
	if(l <= mid) update(l,r,val,LL(idx));
	if(mid < r)	update(l,r,val,RR(idx));
	tt[idx].node = max(tt[LL(idx)].node,tt[RR(idx)].node);
}
 
Node query(int w,int idx) {
	if(tt[idx].node.val < w) {
		return error;
	}
	if(tt[idx].left == tt[idx].right) {
		return tt[idx].node;
	}
	if(tt[LL(idx)].node.val >= w) {
		return query(w,LL(idx));
	} else {
		return query(w,RR(idx));
	}
}
 
int main() {
	error.idx = -1;
	while(scanf("%d%d%d",&n,&h,&m) == 3) {
		checkmin(n,m);
		build(1,n,1);
		while(m --) {
			int w;
			scanf("%d",&w);
			Node ret = query(w,1);
			printf("%d\n",ret.idx);
			if(ret.idx != -1) {
				update(ret.idx,ret.idx,-w,1);
			}
		}
	}
	return 0;
} 

MiYu 2010-09-26 22:15 发表评论

HDOJ 1698 HDU 1698 Just a Hook ACM 1698 IN HDU

MiYu — Fri, 17 Sep 2010 02:09:00 GMT

摘要: MiYu原创, 转帖��h��?: 转蝲�?nbsp;______________白白の屋题目地址 : http://acm.hdu.edu.cn/showproblem.php?pid=1698题目描述 : Just a HookTime Limit: 40... 阅读全文

MiYu 2010-09-17 10:09 发表评论

HDOJ 1754 HDU 1754 I Hate It ACM 1754 IN HDU

MiYu — Wed, 15 Sep 2010 08:08:00 GMT

MiYu原创, 转帖��h��?: 转蝲�?nbsp;______________白白の屋

题目地址:

　　http://acm.hdu.edu.cn/showproblem.php?pid=1754

题目描述:

I Hate It

Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6306 Accepted Submission(s): 2267

Problem Description

很多学校��行一�U�比较的习惯。老师们很喜欢询问�Q�从某某到某某当中，分数最高的是多��?br>�q�让很多学生很反感�?br>
不管你喜不喜�Ƣ，现在需要你做的是，��是按照老师的要求，写一个程序，模拟老师的询问。当�Ӟ��老师有时候需要更新某位同学的成�W�?/div>

Input

本题目包含多�l�测试，请处理到文�g�l�束�?br>在每个测试的�W�一行，有两个正整数 N �?M ( 0学生ID�~�号分别�?�~�到N�?br>�W�二行包含N个整敎ͼ�代表�q�N个学生的初始成�W�Q�其中第i个数代表ID为i的学生的成�W�?br>接下来有M行。每一行有一个字�W?C (只取'Q'�?U') �Q�和两个正整数A�Q�B�?br>当C�?Q'的时候，表示�q�是一条询问操作，它询问ID从A到B(包括A,B)的学生当中，成�W最高的是多��?br>当C�?U'的时候，表示�q�是一条更新操作，要求把ID为A的学生的成�W更改为B�?br>

Output

对于每一�ơ询问操作，在一行里面输出最高成�l��?/div>

Sample Input

5 6 1 2 3 4 5 Q 1 5 U 3 6 Q 3 4 Q 4 5 U 2 9 Q 1 5

Sample Output

5 6 5 9  Hint
Huge input,the C function scanf() will work better than cin

感觉好久没有A题了 , 最�q�一直没有状�? 豆豆也�{行了, 郁闷....... 因�ؓ打算专精数据�l�构斚w��,

所以这几天一直都在复�?数据�l�构, 再一�ơ学习了 �U�段�? 以前只会用它�?更新�?求和 , 现在�l�于水了一

�?RMQ 的裸题了, HAPPY 一�?...

对于 RMQ 的题�? 看PPT 上面�?DP 我直�?rz�?..........表示DP只会做水�?... �q�方面还是交�l?br>

YCH �? 不过看了 shǎ�?/a> 大神博客�?�U�段树专辑后, 发现用线�D�|��处理 �q�类问题非常方便, 修改查询

都是 O (logN)�?, �E�稍优化了下输入, 234MS AC ........

代码如下 :

代码

/*
Coded By  : MiYu
Link      : http://www.cnblogs.com/MiYu  || http://www.shnenglu.com/MiYu
Author By : MiYu
Test      : 1
Program   : 1754
*/
//#pragma warning( disable:4789 )
#include <iostream>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <utility>
#include <queue>
#include <stack>
#include <list>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
using namespace std;
inline int max ( int a, int b ){
    return a > b ? a : b;
}
typedef struct seg_Tree {
    int left, right;
    int mid() { return (left+right)>>1; }
    int max;
}S;
S seg[605000];
int key[200010];
int creat ( int left, int right, int root = 1 ){
    seg[root].left = left;
    seg[root].right = right;
    if ( left == right )
        return seg[root].max = key[left];
    int mid = seg[root].mid();
    return seg[root].max = max ( creat ( left, mid, root << 1 ),creat ( mid + 1, right, ( root << 1 ) + 1 ) );
}

void modify ( int val, int pos, int r = 1 ){
    if ( seg[r].left == seg[r].right ){
        seg[r].max = val;
        return;
    }
    int mid = seg[r].mid();
    if ( pos <= mid ){
        modify ( val, pos, r << 1 );
    } else {
        modify ( val, pos, ( r << 1 ) + 1 );
    }
    seg[r].max = max ( seg[r<<1].max, seg[ (r<<1) + 1 ].max );
}

int quy ( int left, int right, int r = 1 ){
    if ( seg[r].left == left && seg[r].right == right ){
        return seg[r].max;
    }
    int mid = seg[r].mid();
    if ( right <= mid  ){
        return quy ( left, right, r << 1 );
    } else if ( left > mid ) {
        return quy ( left, right, (r << 1) + 1 );
    } else {
        return max ( quy ( left, mid, r << 1 ), quy ( mid + 1, right, (r << 1) + 1 ) );
    }
}
inline bool scan_d(int &num)
{
        char in;bool IsN=false;
        in=getchar();
        if(in==EOF) return false;
        while(in!='-'&&(in<'0'||in>'9')) in=getchar();
        if(in=='-'){ IsN=true;num=0;}
        else num=in-'0';
        while(in=getchar(),in>='0'&&in<='9'){
                num*=10,num+=in-'0';
        }
        if(IsN) num=-num;
        return true;
}
int main ()
{
    int N, M, x, y;
    while ( scan_d(N) && scan_d(M) ){
        for ( int i = 1; i <= N; ++ i ){
            scan_d( key[i] );
        }
        creat ( 1, N );
        while ( M -- ){
            char ask[5];
            scanf ( "%s", ask );
            scan_d(x);
            scan_d(y);
            switch ( ask[0] ){
                case 'Q':    printf ( "%d\n", quy ( x,y ) );
                            break;
                case 'U':    modify ( y, x );
            }
        }
    }
    return 0;
}

/*
5 6
1 2 3 4 5
Q 1 5
U 3 6
Q 3 4
Q 4 5
U 2 9
Q 1 5
*/

MiYu 2010-09-15 16:08 发表评论

HDOJ 1166 HDU 1166 敌兵布阵 (�U�段树解�?ACM 1166 IN HDU

MiYu — Sat, 04 Sep 2010 04:53:00 GMT

MiYu原创, 转帖��h��?: 转蝲�?nbsp;______________白白の屋

先前做了�q�一�? 不过是用树状数组做的, 对于�q�一�c�d��的题目也来说非常方便快捷. 具体地址 : http://www.cnblogs.com/MiYu/archive/2010/08/25/1808441.html

�q�几天学�?�U�段�?, 不是很明白它的用�?�?使用�Ҏ(gu��)��, 因�ؓ 听说树状数组�?�U�段树的一�U�特�D�情�?( 部分资料是这么说�?) . 于是��又做了一�ơ这个题�?

不过使用的是�U�段树的 �Ҏ(gu��)�� :

代码如下 :

代码/*
MiYu原创, 转帖��h���?nbsp;: 转蝲�?nbsp;______________白白の屋
          http://www.cnblog.com/MiYu
Author By : MiYu
Test      : 3
Program   : 1166
*/

#include <iostream>
#include <algorithm>
using namespace std;
typedef struct Line{		//�U�段树结构体
        friend struct optLine;
        Line () { sum = 0; left = right = NULL; };
        int l,r,sum;
        struct Line *left, *right;
}L;
typedef struct optLine {	//�U�段树操作结�?br>       void Creat ( struct Line *node, int beg, int end );
       void modify ( struct Line *node, int pos,int val );
       int quy ( struct Line *node, int beg, int end );     
}OPT;
void optLine::Creat ( struct Line *node, int beg, int end ){	//建树
     node->l = beg;  node->r = end;
     if ( beg == end ) return;
     int mid = ( node->l + node->r ) >> 1;
     L *p = new L;  L *q = new L;
     node->left = p; node->right = q;
     Creat ( p, beg, mid );   Creat ( q, mid + 1, end );          
}
void optLine::modify ( struct Line *node, int pos, int val ){	//修改
     if ( pos >= node->l && pos <= node->r ) node->sum += val;
     if ( node->l == node->r ) return;
     int mid = ( node->l + node->r ) >> 1;
     if ( mid >= pos ) modify ( node->left, pos, val );
     else if ( mid < pos ) modify ( node->right, pos,val );
}
int optLine::quy ( struct Line *node, int beg, int end ){	//查询
    int sum = 0;  int mid = ( node->l + node->r ) >> 1;
    if ( node->l == beg && node->r == end ) return node->sum;
    if ( mid >= end ) return sum += quy ( node->left, beg,end );
    else if ( beg > mid ) return sum += quy ( node->right, beg, end ); 
    else {
          return sum += quy ( node->left, beg, mid ) + quy ( node->right, mid+1,end );     
    }   
}
inline bool scan_ud(int &num) 
{
        char in;
        in=getchar();
        if(in==EOF) return false;
        while(in<'0'||in>'9') in=getchar();
        num=in-'0';
        while(in=getchar(),in>='0'&&in<='9'){
                num*=10,num+=in-'0';
        }
        return true;
}
int main ( int T, int N )
{
    scan_ud ( T );
    for ( int ca = 1; ca <= T ; ++ ca ){
         printf ( "Case %d:\n",ca );
         scan_ud ( N );    OPT opt;  L *root = new L;   int val,x,y;
         opt.Creat ( root, 1, N );  
         for ( int i = 1; i <= N; ++ i ){
              scan_ud ( val );  opt.modify ( root, i, val );    
         }
         char ask[10];  
         while ( scanf ( "%s", ask ), ask[0] != 'E' ){
                scanf ( "%d%d",&x,&y );
                switch ( ask[0] ){
                        case 'Q':  printf ( "%d\n",opt.quy ( root, x,y ) ); break;
                        case 'A':  opt.modify ( root, x, y ); break;
                        case 'S':  opt.modify ( root, x, -y ); break;      
                }      
         }
    }
    return 0;
}

MiYu 2010-09-04 12:53 发表评论

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14814		Accepted: 6404

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HDU 3468 HDOJ 3468 A Simple Problem with Integers ACM 3468 IN HDU

PKU 2528 POJ 2528 Mayor's posters ( �U�段�?���L���?) ACM 2528 IN PKU

PKU 2352 POJ 2352 Stars ( �U�段树版 ) ACM 2352 IN PKU

HDU 3016 HDOJ 3016 Man Down ACM 3016 IN HDU

Man Down

HDU 2871 HDOJ 2871 Memory Control ACM 2871 IN HDU

PKU 3667 HDOJ 3667 Hotel ACM 3667 IN HDU

HDU 1540 HDOJ 1540 Tunnel Warfare ACM 1540 IN HDU

�U�段树的一�U�简化实现[转] by �t�雪赤兔

HDU 2795 HDOJ 2795 Billboard ACM 2795 IN HDU

Billboard

HDOJ 1698 HDU 1698 Just a Hook ACM 1698 IN HDU

HDOJ 1754 HDU 1754 I Hate It ACM 1754 IN HDU

I Hate It

HDOJ 1166 HDU 1166 敌兵布阵 (�U�段树解�?ACM 1166 IN HDU

PKU 2528 POJ 2528 Mayor's posters ( �U�段�?��L��?) ACM 2528 IN PKU