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…

题目链接:HDU6214 留一个链式前向星+Dinic模板(希望不要被某人发现,嘿嘿嘿). #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <queue> using namespace std; #define next Next const int inf = 0x3f3f3f3f; ; int level[maxn];…

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…

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…

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…

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…

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…

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…

题目链接 ISAP写法 #include <bits/stdc++.h> using namespace std; typedef long long LL; namespace FastIO { const static int MX=1e6; ; char nc() { static char buf[MX],*p1=buf+MX,*pend=buf+MX; if(p1==pend) { p1=buf; pend=buf+fread(buf,,MX,stdin); if(pend==p1)…

先最大流跑一遍 在残存网络上把满流边容量+1 非满流边容量设为无穷大 在进行一次最大流即可 (这里的边都不包括建图时用于反悔的反向边) #include<cstdio> #include<cstring> #include<cstdlib> #include<queue> #include<iostream> #define rep(i,a,b) for(int i=a;i<=b;++i) using namespace std; ; in…