POJ2914 Minimum Cut —— 最小割

题目链接：http://poj.org/problem?id=2914

Minimum Cut

Time Limit: 10000MS Memory Limit: 65536KTotal Submissions: 10117 Accepted: 4226Case 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 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)

代码如下：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <cmath>#include <queue>#include <stack>#include <map>#include <string>#include <set>using namespace std;typedef long long LL;const int INF = 2e9;const LL LNF = 9e18;const int mod = 1e9+7;const int MAXN = 500+10;int mp[MAXN][MAXN];bool combine[MAXN];int n, m;bool vis[MAXN];int w[MAXN];int prim(int times, int &s, int &t) //最大生成树？{    memset(w,0,sizeof(w));    memset(vis,0,sizeof(vis));    for(int i = 1; i<=times; i++)   //times为实际的顶点个数    {        int k, maxx = -INF;        for(int j = 0; j<n; j++)            if(!vis[j] && !combine[j] && w[j]>maxx)                maxx = w[k=j];        vis[k] = 1;        s = t; t = k;        for(int j = 0; j<n; j++)            if(!vis[j] && !combine[j])                w[j] += mp[k][j];    }    return w[t];}int mincut(){    int ans = INF;    memset(combine,0,sizeof(combine));    for(int i = n; i>=2; i--)   //每一次循环，就减少一个点    {        int s, t;        int tmp = prim(i, s, t);        ans = min(ans, tmp);        combine[t] = 1;        for(int j = 0; j<n; j++)   //把t点删掉，与t相连的边并入s        {            mp[s][j] += mp[t][j];            mp[j][s] += mp[j][t];        }    }    return ans;}int main(){    while(scanf("%d%d",&n,&m)!=EOF)    {        memset(mp,0,sizeof(mp));        for(int i = 1; i<=m; i++)        {            int u, v, w;            scanf("%d%d%d",&u,&v,&w);            mp[u][v] += w;            mp[v][u] += w;        }        cout<< mincut() <<endl;    }    return 0;}