POJ2914 Minimum Cut(最小割模板题)

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本文介绍了一种解决无向图中最小割问题的方法。通过多次搜索来逐步减少顶点数并合并相邻顶点，最终找到使得图断开所需的最少边数。输入包括顶点数、边数及连接信息，输出为最小割大小。

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Minimum Cut

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 9122		Accepted: 3820
Case 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

Sample Output

2
1
2#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define maxn 1005
#define inf 1000000007
int mp[maxn][maxn];
int N, M;
bool combine[maxn];
int minC = inf;
void Search(int &s, int &t){
    bool vis[maxn];
    int w[maxn];
    memset(vis, 0, sizeof vis);
    memset(w, 0, sizeof w);
    int tmpj;
    for(int i = 0;i < N;i++){
        int MAX = -inf;
        for(int j = 0;j < N;j++){
            if(!vis[j]&&!combine[j]&&MAX<w[j]){
                MAX = w[j];
                tmpj = j;
            }
        }
        if(t==tmpj) {
            minC=w[t];
            return ;
        }
        vis[tmpj] = 1;
        s = t, t = tmpj;
        for(int j = 0;j < N;j++){
            if(!vis[j]&&!combine[j])
                w[j]+=mp[t][j];
        }
    }
    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;
        Search(s, t);
        combine[t] = true;
        ans=ans>minC?minC:ans;
        for(int j = 0;j < N;j++){
            mp[s][j] += mp[t][j];
            mp[j][s] += mp[j][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());
    }
}