Drainage Ditches Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 16314    Accepted Submission(s): 7748 Problem Description Every time it rains on Farmer John's fields, a pond forms over Bessie'…

Drainage Ditches Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 62016 Accepted: 23808 Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by wate…

Drainage Ditches Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8861    Accepted Submission(s): 4141 Problem Description Every time it rains on Farmer John's fields, a pond forms over Bessie's…

使用stringstream对象简化类型转换 C++标准库中的<sstream>提供了比ANSI C的<stdio.h>更高级的一些功能,即单纯性.类型安全和可扩展性.在本文中,我将展示怎样使用这些库来实现安全和自动的类型转换. 为什么要学习 如果你已习惯了<stdio.h>风格的转换,也许你首先会问:为什么要花额外的精力来学习基于<sstream>的类型转换呢?也许对下面一个简单的例子的回顾能够说服你.假设你想用sprintf()函数将一个变量从int类型…

Going Home Description On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertically, to an adjacent point. For each little man, you need to pay a $1 travel fee for e…

#include <cstdio> #include <iostream> #include <cstring> #include<queue> #include<cmath> using namespace std; const int INF = 0x3fffffff; int g[1005][1005]; int pre[1005]; int m; int bfs(int s,int t) { queue<int>q; q.pu…

Drainage Ditches Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 65146 Accepted: 25112 Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by wate…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1532 Drainage Ditches Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 18864    Accepted Submission(s): 8980 Problem Description Every time it rain…

#include<cstdio> #include<cstring> #include<algorithm> using namespace std; int n, m, S, T; ;//点数的最大值 ;//边数的最大值 const int INF = 0x3f3f3f3f; struct Edge { int to,next,cap,flow; } edge[MAXM]; //注意是MAXM int tol; int head[MAXN]; void init()…

http://poj.org/problem?id=1273 Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer Jo…

题目链接:http://poj.org/problem?id=1273 Time Limit: 1000MS Memory Limit: 10000K Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes…

pid=1532">Drainage Ditches Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8574    Accepted Submission(s): 3991 Problem Description Every time it rains on Farmer John's fields, a pond forms…

题目链接: Drainage Ditches Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 67475   Accepted: 26075 Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is cover…

题目链接:poj1273 Drainage Ditches 呜呜,今天自学网络流,看了EK算法,学的晕晕的,留个简单模板题来作纪念... #include<cstdio> #include<vector> #include<queue> #include<cstring> #include<set> #include<algorithm> #define CLR(a,b) memset((a),(b),sizeof((a))) usi…

题目链接:http://poj.org/problem?id=1273 Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has…

Drainage Ditches Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9715    Accepted Submission(s): 4623 Problem Description Every time it rains on Farmer John's fields, a pond forms over Bessie's…

Drainage Ditches 自己拉的专题里面没有这题,网上找博客学习网络流的时候看到闯亮学长的博客然后看到这个网络流入门题!随手一敲WA了几发看讨论区才发现坑点! 本题采用的是Edmonds-Karp算法求增广路.小白书上只介绍了这个算法,确实对于数据不刁钻的题目这个算法足以应对.大白书上的Dinic及SAP.ISAP还没有去看,以后慢慢攻克吧! 回到这个题:n条水渠,m个交点,每条水渠有一个最大容量cap,求1号交点到m号交点的最大流. 学了EK算法这题就可以裸引用了,只是注意一个坑点:…

Drainage Ditches Problem Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has…

Drainage Ditches Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9580    Accepted Submission(s): 4541 Problem Description Every time it rains on Farmer John's fields, a pond forms over Bessie's…

Drainage DitchesHal Burch Time Limit 1000 ms Memory Limit 65536 kb description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a…

Drainage Ditches Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17065    Accepted Submission(s): 8066 Problem Description Every time it rains on Farmer John's fields, a pond forms over Bessie'…

描述 Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches s…

                                                                                        Drainage Ditches   Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water f…

#include<cstdio> #include<algorithm> #include<cstring> #include<queue> #define N 250 #define M 250 #define INF 100000000 using namespace std; int head[N],cur[N],lev[N],ecnt=1,S,T,n,m; queue <int> q; struct adj { int nxt,v,w;…

网络流一直没学,来学一波网络流. https://vjudge.net/problem/POJ-1273 题意:给定点数,边数,源点,汇点,每条边容量,求最大流. 解法:EK或dinic. EK:每次增广用bfs选择一条从源到汇具有最少边数的增广路径,然后找出该路径容量最小的边,就是此次增加的流量,然后沿该路径增加反向边,同时修改每条边的容量,重复上述过程直到找不到增广路(即minFlow = 0)为止. dinic: 每次bfs从源点到汇点分层(层数是源点到它最少要经过的边数),然后dfs从源…

题目大意:最大流的模板题...源点是0,汇点是n-1. 代码如下: # include<iostream> # include<cstdio> # include<cmath> # include<string> # include<vector> # include<list> # include<set> # include<map> # include<queue> # include<…

解题关键:最大流裸题 #include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<cmath> #include<iostream> #include<queue> #include<vector> #define inf 0x3f3f3f3f #define MAX_V 252 using namespac…

这道题用dinic会超时 用E_K就没问题 注意输入数据有重边.POJ1273 dinic的复杂度为O(N*N*M)E_K的复杂度为O(N*M*M)对于这道题,复杂度是相同的. 然而dinic主要依靠Dfs寻找增广路,故而使用了太多次递归,而利用bfs寻找增广路(使用队列而不用递归)的EK等于用栈的方式实现了dinic的递归,所以 大幅提高了效率. 最大流算法的核心就是找到增广路并且增加反向边流量,理解了这个,最大流算法就很简单了. #include<iostream> #include<…

POJ 1273给出M条边,N个点,求源点1到汇点N的最大流量. 本文主要就是附上dinic的模板,供以后参考. #include <iostream> #include <stdio.h> #include <algorithm> #include <queue> #include <string.h> /* POJ 1273 dinic算法模板 边是有向的,而且存在重边,且这里重边不是取MAX,而是累加和 */ using namespace…

http://poj.org/problem?id=1273 题目大意: n点m边网络流,求1-n最大流. —————————————— 网络流板子,切了. #include <cstdio> #include <iostream> #include <cmath> #include <cstring> #include <algorithm> using namespace std; ; ; inline int read(){ ,w=;; '…