POJ2914 Minimum Cut（最小割）

Minimum Cut Time Limit: 10000MS Memory Limit: 65536KTotal Submissions: 9715 Accepted: 4085Case Time Limit: 5000MS

Description

Given an undirected graph, in which two vertices can be connected by multiple edges, what is the size of the minimum cut of the graph? i.e. how many edges must be removed at least to disconnect the graph into two subgraphs?

Input

Input contains multiple test cases. Each test case starts with two integers N and M (2 ≤ N ≤ 500, 0 ≤ M ≤ N × (N − 1) ⁄ 2) in one line, where N is the number of vertices. Following are M lines, each line contains M integers A, B and C (0 ≤ A, B < N, A ≠ B, C > 0), meaning that there C edges connecting vertices A and B.

Output

There is only one line for each test case, which contains the size of the minimum cut of the graph. If the graph is disconnected, print 0.

Sample Input

3 3 0 1 1 1 2 1 2 0 1 4 3 0 1 1 1 2 1 2 3 1 8 14 0 1 1 0 2 1 0 3 1 1 2 1 1 3 1 2 3 1 4 5 1 4 6 1 4 7 1 5 6 1 5 7 1 6 7 1 4 0 1 7 3 1

Sample Output

2 1 2

Source

Baidu Star 2006 Semifinal  Wang, Ying (Originator)  Chen, Shixi (Test cases) ———————————————————————————————————

题目的意思是给出一张图，求整体最小割

思路：模板题

#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <algorithm> #include <cmath> #include <map> #include <set> #include <stack> #include <queue> #include <vector> #include <bitset> using namespace std; #define LL long long const int INF = 0x3f3f3f3f; #define MAXN 800 int mp[MAXN][MAXN],w[MAXN],vis[MAXN],combine[MAXN]; int minC,cnt,n,m; void Find(int &s,int &t) { memset(vis,0,sizeof(vis)); memset(w,0,sizeof(w)); int tmp=INF; int q=n; while(q--) { int mx=-INF; for(int i=0; i<n; i++) { if(!vis[i]&&!combine[i]&&mx<w[i]) { mx=w[i]; tmp=i; } } if(t==tmp) { minC=w[t]; return; } vis[tmp]=1; s=t,t=tmp; for(int i=0;i<n;i++) { if(!vis[i]&&!combine[i]) w[i]+=mp[t][i]; } } minC=w[t]; } int mincut() { int ans=INF; int s,t; memset(combine,0,sizeof(combine)); for(int i=0; i<n-1; i++) { s=t=-1; Find(s,t); combine[t]=true; ans=min(ans,minC); for(int i=0;i<n;i++) { mp[s][i]+=mp[t][i]; mp[i][s]+=mp[i][t]; } } return ans; } int main() { while(~scanf("%d%d",&n,&m)) { memset(mp,0,sizeof(mp)); int u,v,w; while(m--) { scanf("%d %d %d",&u,&v,&w); mp[u][v]+=w; mp[v][u]+=w; } printf("%d\n",mincut()); } return 0; }