题意：。。。。emm。。。就是一个最小割最大流，。，。。。用dinic跑一遍。。
然后让你输出割边，就是 u为能从起点到达的点， v为不能从起点到达的点
最后在残余路径中用dfs跑一遍 能到达的路标记一下
然后循环判断输出即可 还有不要忘了是正向路 所以循环时i+=2
#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <set> #include <vector> #include <stack> #include <queue> #include <algorithm> #include <cmath> #define MOD 2018 #define LL long long #define ULL unsigned long long #define maxn 100009 #define Pair pair<int, int> #define mem(a, b) memset(a, b, sizeof(a)) #define _ ios_base::sync_with_stdio(0),cin.tie(0) //freopen("1.txt", "r", stdin); using namespace std; const int INF = 0x3f3f3f3f; int head[maxn], d[maxn], cur[maxn], vis[maxn]; int n, m, s, t; int cnt = 0; struct node{ int u, v, c, next; }Node[maxn*2]; void add_(int u, int v, int c) { Node[cnt].u = u; Node[cnt].v = v; Node[cnt].c = c; Node[cnt].next = head[u]; head[u] = cnt++; } void add(int u, int v, int c) { add_(u, v, c); add_(v, u, c); } bool bfs() { queue<int> Q; mem(d,0); Q.push(s); d[s] = 1; while(!Q.empty()) { int u = Q.front(); Q.pop(); for(int i=head[u]; i!=-1; i=Node[i].next) { node e = Node[i]; if(!d[e.v] && e.c > 0) { d[e.v] = d[e.u] + 1; Q.push(e.v); if(e.v == t) return 1; } } } return d[t] != 0; } int dfs(int u, int cap) { if(u == t || cap == 0) return cap; int ret = 0; for(int &i=cur[u]; i!=-1; i=Node[i].next) { node e = Node[i]; if(d[e.v] == d[e.u] + 1 && e.c > 0) { int V = dfs(e.v, min(cap, e.c)); Node[i].c -= V; Node[i^1].c += V; cap -= V; ret += V; if(cap == 0) break; } } return ret; } int dinic() { int ans = 0; while(bfs()) { memcpy(cur, head, sizeof(head)); ans += dfs(s, INF); } return ans; } void find(int u) { for(int i=head[u]; i!=-1; i=Node[i].next) { node e = Node[i]; if(vis[e.v] || e.c == 0) continue; vis[e.v] = 1; find(e.v); } } int main() { while(~scanf("%d%d",&n,&m) && n+m) { mem(vis,0); mem(head,-1); cnt = 0; s = 1; t = 2; for(int i=0; i<m; i++) { int u, v, c; scanf("%d%d%d",&u,&v,&c); add(u, v, c); } dinic(); vis[s] = 1; find(s); //寻找能从s到达的路 for(int i=0; i<cnt; i+=2) if(vis[Node[i].u] && !vis[Node[i].v] || !vis[Node[i].u] && vis[Node[i].v]) // 如果一个能到达 另一个不能到达 则为割边 输出即可 cout<< Node[i].u << " " << Node[i].v <<endl; cout<< endl; } return 0; }