題意：輸入一個n和m，n為點數，m為邊數，接下來m行u， v， w，求1和2不連線的最小花費。

解題：最大流最小割定理，要輸出割斷的邊。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;
struct node {
    int from, to, cap, next;
} edge[2000000];
int dep[100000], head[100000], vist[1000000];
int src, des, cnt, n, m;
const int INF = 1e9;

void addedge(int u, int v, int w)
{
    edge[cnt].from = u, edge[cnt].to = v, edge[cnt].cap = w, edge[cnt].next = head[u];
    head[u] = cnt++;

    edge[cnt].from = v, edge[cnt].to = u, edge[cnt].cap = w, edge[cnt].next = head[v];
    head[v] = cnt++;
    return ;
}

int bfs()
{
    queue<int>que;
    memset(dep, -1, sizeof(dep));
    dep[src] = 0;
    que.push(src);
    while(!que.empty()) {
        int u = que.front();
        que.pop();
        for(int i = head[u]; i != -1; i = edge[i].next) {
//            printf("**********\n");
            int v = edge[i].to;
            if(dep[v]==-1 && edge[i].cap > 0) {
                dep[v] = dep[u]+1;
                que.push(v);
            }
        }
    }
    return dep[des]!=-1;
}

int dfs(int u, int minc)
{
    if(u == des) return minc;
    int tmp;
    for(int i = head[u]; i != -1; i = edge[i].next) {
        int v = edge[i].to;
        if(dep[v]==dep[u]+1 && edge[i].cap>0 && (tmp=dfs(v, min(minc, edge[i].cap))) ) {
            edge[i].cap -= tmp;
            edge[i^1].cap += tmp;
            return tmp;
        }
    }
    dep[u] = -1;
    return 0;
}

int dinic()
{
    int ans = 0, tmp;
    while(bfs()) {
        while(tmp=dfs(src, INF)) ans += tmp;
    }
    return ans;
}

void bbfs(int s, int c) // 記錄點集
{
    queue<int> que;
    que.push(s);
    vist[s] = c;
    while(!que.empty()) {
        int u = que.front();
        que.pop();
        for(int i = head[u]; i!=-1; i = edge[i].next) {
            int v = edge[i].to;
            if(edge[i].cap > 0 && !vist[v]) {
                vist[v] = c;
                que.push(v);
            }
        }
    }
    return ;
}

int main()
{
    int u, v, w;
    while(~scanf("%d%d", &n, &m) && (n+m)) {
        src = 1, des = 2;
        cnt = 0;
        memset(head, -1, sizeof(head));
        while(m--) {
            scanf("%d%d%d", &u, &v, &w);
            addedge(u, v, w);
        }

        dinic();
        memset(vist, 0, sizeof(vist));
        bbfs(src, 1);
        bbfs(des, 2);
        for(int i = 0; i < cnt; i++)
        {
            if(edge[i].cap==0 && vist[edge[i].from]!=vist[edge[i].to])
            printf("%d %d\n", edge[i].from, edge[i].to);
        }
        printf("\n");
    }
    return 0;
}