UVA-10480 Sabotage(最小割，割边方案)

题目链接：UVA-10480 Sabotage

题意

给出一个图的连边(无向边)和割断每条边的花费，要割掉一些边令结点1和结点2不连通，求最小花费的切割方案。

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思路

在原图上以结点1为源点，结点2为汇点，边的花费为容量建立流网络，最小割即为最小花费。

设割边将流网络的结点分为$S$和$T$两个集合。Ford-Fulkerson方法求最大流后的残量网络中，从源点开始dfs，每次走残余容量大于$0$的边，即可找到所有$S$集合内的结点。所有边$(u,v)$中，若$u\in S, v\in T$，则$(u,v)$为割边。

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代码实现

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using std::queue;
const int INF = 0x3f3f3f3f, N = 60, M = 1100;
int head[N], d[N];
int s, t, tot, maxflow;
bool vis[N];
struct Edge
{
    int to, cap, nex;
} edge[M];
queue<int> q;
void add(int x, int y, int z) {
    edge[++tot].to = y, edge[tot].cap = z, edge[tot].nex = head[x], head[x] = tot;
}
bool bfs() {
    memset(d, 0, sizeof(d));
    while (q.size()) q.pop();
    q.push(s); d[s] = 1;
    while (q.size()) {
        int x = q.front(); q.pop();
        for (int i = head[x]; i; i = edge[i].nex) {
            int v = edge[i].to;
            if (edge[i].cap && !d[v]) {
                q.push(v);
                d[v] = d[x] + 1;
                if (v == t) return true;
            }
        }
    }
    return false;
}
int dinic(int x, int flow) {
    if (x == t) return flow;
    int rest = flow, k;
    for (int i = head[x]; i && rest; i = edge[i].nex) {
        int v = edge[i].to;
        if (edge[i].cap && d[v] == d[x] + 1) {
            k = dinic(v, std::min(rest, edge[i].cap));
            if (!k) d[v] = 0;
            edge[i].cap -= k;
            edge[i^1].cap += k;
            rest -= k;
        }
    }
    return flow - rest;
}
void init() {
    tot = 1, maxflow = 0;
    s = 1, t = 2;
    memset(head, 0, sizeof(head));
    memset(vis, false, sizeof(vis));
}
void dfs(int u) {
  vis[u] = true;
  for (int i = head[u]; i; i = edge[i].nex) {
    int v = edge[i].to;
    if (!vis[v] && edge[i].cap) dfs(v);
  }
}

int main() {
    int n, m;
    while (~scanf("%d %d", &n, &m) && (n || m)) {
        init();
        for (int i = 0, u, v, z; i < m; i++) {
            scanf("%d %d %d", &u, &v, &z);
            add(u, v, z);
            add(v, u, z);
        }
        while (bfs()) maxflow += dinic(s, INF);
        dfs(s);
        for (int i = 1; i <= n; i++) {
            if (vis[i]) {
                for (int j = head[i]; j; j = edge[j].nex) {
                    if (!vis[edge[j].to]) printf("%d %d\n", i, edge[j].to);
                }
            }
        }
        puts("");
    }
    return 0;
}

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