标签：poj2914
| Time Limit: 10000MS | Memory Limit: 65536K | |
| Total Submissions: 7610 | Accepted: 3203 | |
| 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 integersA, 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
#include <stdio.h>
#include <string.h>
#define inf 0x7fffffff
#define maxn 502
int map[maxn][maxn], W[maxn], hash[maxn];
bool vis[maxn];
void getMap(int n, int m)
{
int i, u, v, c;
memset(map, 0, sizeof(map));
for(i = 0; i < m; ++i){
scanf("%d%d%d", &u, &v, &c);
map[u][v] += c; map[v][u] += c;
}
}
int Stoer_Wagner(int n)
{
int minCut = inf, nowCut, now, pre, i, j;
for(i = 0; i < n; ++i) hash[i] = i;
while(n > 1){
nowCut = -1; now = 1; vis[hash[0]] = 1; pre = 0;
for(i = 1; i < n; ++i){
W[hash[i]] = map[hash[0]][hash[i]];
vis[hash[i]] = 0;
if(W[hash[i]] > nowCut){
nowCut = W[hash[i]]; now = i;
}
}
for(j = 1; j < n; ++j){
vis[hash[now]] = 1;
if(j == n - 1){
if(nowCut < minCut) minCut = nowCut;
for(i = 0; i < n; ++i){
map[hash[pre]][hash[i]] += map[hash[now]][hash[i]];
map[hash[i]][hash[pre]] += map[hash[now]][hash[i]];
}
hash[now] = hash[--n];
break;
}
pre = now; nowCut = -1;
for(i = 1; i < n; ++i){
if(!vis[hash[i]]){
W[hash[i]] += map[hash[pre]][hash[i]];
if(W[hash[i]] > nowCut){
nowCut = W[hash[i]]; now = i;
}
}
}
}
}
return minCut;
}
void solve(int n)
{
printf("%d\n", Stoer_Wagner(n));
}
int main()
{
freopen("stdin.txt", "r", stdin);
int n, m;
while(scanf("%d%d", &n, &m) == 2){
getMap(n, m);
solve(n);
}
return 0;
}POJ2914 Minimum Cut 【全局最小割】（Stoer_Wagner）
标签：poj2914
原文地址：http://blog.csdn.net/chang_mu/article/details/38928923
