Smallest Minimum Cut Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 2281    Accepted Submission(s): 913 Problem Description Consider a network G=(V,E) with source s and sink t. An s-t cut is a…

Problem Description Consider a network G=(V,E) with source s and sink t. An s-t cut is a partition of nodes set V into two parts such that s and t belong to different parts. The cut set is the subset of E with all edges connecting nodes in different…

Problem Description Consider a network G=(V,E) with source s and sink t . An s-t cut is a partition of nodes set V into two parts such that s and t belong to different parts. The cut set is the subset of E with all edges connecting nodes in different…

Smallest Minimum Cut Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 1181    Accepted Submission(s): 473 Problem Description Consider a network G=(V,E) with source s and sink t. An s-t cut is a…

hdu 6214 Smallest Minimum Cut[最大流] 题意:求最小割中最少的边数. 题解:对边权乘个比边大点的数比如300,再加1 ,最后,最大流对300取余就是边数啦.. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<vector> #include<queue…

Consider a network G=(V,E) G=(V,E) with source s s and sink t t . An s-t cut is a partition of nodes set V V into two parts such that s s and t t belong to different parts. The cut set is the subset of E E with all edges connecting nodes in different…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6214 Problem Description Consider a network G=(V,E) with source s and sink t. An s-t cut is a partition of nodes set V into two parts such that s and t belong to different parts. The cut set is the subse…

双倍经验题:HDU 6214,3987 求最小割的最小边. 方案一: 首先跑最大流,这个时候割上都满载了,于是将满载的边 cap = 1,其他 inf ,再跑最大流,这个时候限定这个网络的关键边就是那个最少边的那个割. 方案二: 奇技淫巧,将每条边 cap* A + 1,最大流 = flow / A ,最小割边 = fow % A: 原理:每条边容量扩大 到 cap * A + 1,那么最大流也一定扩大到 *A + 1,原图是多解,但是新图, 例如最少边的个是2条边,那么他就扩大到了 *A +…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6214 题意:求边数最小的割. 解法: 建边的时候每条边权 w = w * (E + 1) + 1; 这样得到最大流 maxflow / (E + 1) ,最少割边数 maxflow % (E + 1) 道理很简单,如果原先两类割边都是最小割,那么求出的最大流相等 但边权变换后只有边数小的才是最小割了 乘(E+1)是为了保证边数叠加后依然是余数,不至于影响求最小割的结果 因为假设最小割=k,那么现在新…

题意:给定上一个有向图,求 s - t 的最小割且边数最少. 析:设边的容量是w,边数为m,只要把每边打容量变成 w * (m+1) + 1,然后跑一个最大流,最大流%(m+1),就是答案. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <c…