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 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 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): 17065    Accepted Submission(s): 8066 Problem Description Every time it rains on Farmer John's fields, a pond forms over Bessie'…

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 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:…

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…

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<…

最大流模板题 今天补最大流,先写道模板题,顺便写点对它的理解 最大流问题就是给一个幽香有向图,每一条边有容量,问若从$s$点放水,最多会有多少水流到$t$ 为了解决整个问题,第一步我们当然要找到一条路径让它达到满流的状态,这里直接xjbdfs即可 比如我们找到一条路径$1\to2\to3\to4$,流量为$2$ 但是我们有可能需要“反悔”,比如不走$2\to3$而走$2\to4$ 这时我们可以使用一些小技巧:添加反向边并更新正向边的容量 以后如果我们走了反向边就相当于是在“反悔” 我们把上述找一…

http://poj.org/problem?id=1273 Drainage Ditches Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 62708   Accepted: 24150 Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means…