POJ 2914 无向图的最小割，模板题。

Minimum Cut

Time Limit: 10000MS Memory Limit: 65536KTotal Submissions: 9126 Accepted: 3823Case 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, whereN is the number of vertices. Following are M lines, each line containsM integers A, B and C (0 ≤ A, B <N, A ≠ B, C > 0), meaning that there C edges connecting verticesA 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 30 1 11 2 12 0 14 30 1 11 2 12 3 18 140 1 10 2 10 3 11 2 11 3 12 3 14 5 14 6 14 7 15 6 15 7 16 7 14 0 17 3 1

Sample Output

212

Source

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

#include <iostream>#include <cstdio>#include <cstring>#define MAXN 505#define INF 1000000000using namespace std;int map[MAXN][MAXN];int v[MAXN], dis[MAXN]; //v数组是马甲数组，dis数组用来表示该点与A集合中所有点之间的边的长度之和bool vis[MAXN];//用来标记是否该点加入了A集合int Stoer_Wagner(int n){    int i, j, res = INF;    for(i = 0; i < n; i ++)        v[i] = i; //初始马甲为自己    while(n > 1)    {        int k, pre = 0;  //pre用来表示之前加入A集合的点，我们每次都以0点为第一个加入A集合的点        memset(vis, 0, sizeof(vis));        memset(dis, 0, sizeof(dis));        for(i = 1; i < n; i ++)        {            k = -1;            for(j = 1; j < n; j ++)  //根据之前加入的点，要更新dis数组，并找到最大的dis                if(!vis[v[j]])                {                    dis[v[j]] += map[v[pre]][v[j]];                    if(k == -1 || dis[v[k]] < dis[v[j]])                        k = j;                }            vis[v[k]] = true;//标记该点已经加入A集合            if(i == n - 1) //最后一次加入的点就要更新答案了            {                res = min(res, dis[v[k]]);                for(j = 0; j < n; j ++) //将该点合并到pre上，相应的边权就要合并                {                    map[v[pre]][v[j]] += map[v[j]][v[k]];                    map[v[j]][v[pre]] += map[v[j]][v[k]];                }                v[k] = v[-- n];//删除最后一个点            }            pre = k;        }    }    return res;}int main(){    int n, m, u, v, w;    while(scanf("%d%d", &n, &m) != EOF)    {        memset(map, 0, sizeof(map));        while(m --)        {            scanf("%d%d%d", &u, &v, &w);            map[u][v] += w;            map[v][u] += w;        }        printf("%d\n", Stoer_Wagner(n));    }    return 0;}