求无向图的最小割
刷UOJ用户群的时候看到有人问一道全局最小割的裸题。。忽然想起之前还坑着Stoer Wagner算法。。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define FOR(i,j,k) for(i=j;i<=k;++i)
const int N = 501;
int vis[N], dis[N], v[N], a[N][N];
int stoer_wagner(int n) {
int i, j, ans = 0x7fffffff, p, last;
FOR(i,1,n) v[i] = i;
while (n > 1) {
p = last = 0;
FOR(i,2,n) {
dis[v[i]] = a[v[1]][v[i]];
if (dis[v[i]] > dis[v[p]]) p = i;
}
FOR(i,2,n) vis[v[i]] = 0;
vis[v[1]] = 1;
FOR(i,2,n) {
if (i == n) {
ans = min(ans, dis[v[p]]);
FOR(j,1,n) a[v[j]][v[last]] = a[v[last]][v[j]] += a[v[j]][v[p]];
v[p] = v[n--];
}
vis[v[last = p]] = 1; p = -1;
FOR(j,2,n) if (!vis[v[j]]) {
dis[v[j]] += a[v[last]][v[j]];
if (p == -1 || dis[v[p]] < dis[v[j]])
p = j;
}
}
}
return ans;
}
int main() {
int n, m, u, v, w;
while (scanf("%d%d", &n, &m) != EOF) {
memset(a, 0, sizeof a);
while (m--) {
scanf("%d%d%d", &u, &v, &w);
++u; ++v;
a[u][v] += w; a[v][u] += w;
}
printf("%d\n", stoer_wagner(n));
}
return 0;
}
Minimum Cut
Time Limit: 10000MS Memory Limit: 65536K
Total Submissions: 8588 Accepted: 3617
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
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
本文介绍了一种求解无向图全局最小割的问题,并提供了一个使用Stoer-Wagner算法的具体实现。该算法通过逐步合并顶点来寻找图中最小的边割集。
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