题意:
裸的最大流,求最小割
思路:
裸的最大流,要求输出最小割割的边,
首先要了解 「在网络中,如果f是一个流,CUT (S,T)是一个割,且f的值等于割CUT(S,T)的容量,那么f是一个最大流, CUT(S,T)是一个最小割。」(来自百度百科)
所以最小割割的就是满流的最大流的边。
那么在跑完dinic后,如果发现一条线的一段可分层,而另一段不可分层,那么这条边就是满流边,输出即可。
#include <iostream>
#include <string.h>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <stdio.h>
#include <deque>
using namespace std;
#define ll long long
#define ull unsigned long long
#define INF 0x7fffffff
#define maxn 800005
#define eps 0.00000001
#define PI acos(-1.0)
#define M 1000000007
struct Edge{
int v, w, nxt;
}edge[maxn];
int sx, ex, si, ei, x, y;
int pre[100005], tot, dis[100005], cur[100005];
int n, m;
int vis[maxn], a[maxn], b[maxn];
queue<int> que;
void init() {
tot = 0;
memset(pre, -1, sizeof(pre));
memset(vis, 0, sizeof(vis));
si = 1;
ei = 2;
}
void addEdge(int u, int v, int w) {
edge[tot].v = v;
edge[tot].w = w;
edge[tot].nxt = pre[u];
pre[u] = tot ++;
}
int bfs() {
for (int i = 0; i <= n; i ++)
dis[i] = -1;
dis[ei] = 0;
que.push(ei);
while(!que.empty()) {
int u = que.front(); que.pop();
for (int i = pre[u]; i + 1; i = edge[i].nxt) {
if(dis[edge[i].v] == -1 && edge[i ^ 1].w > 0) {
dis[edge[i].v] = dis[u] + 1;
que.push(edge[i].v);
}
}
}
return dis[si] != -1;
}
int dfs(int u, int flow) {
if(u == ei) return flow;
int delte = flow, d;
for (int &i = cur[u]; i + 1; i = edge[i].nxt) {
if(dis[u] == dis[edge[i].v] + 1 && edge[i].w > 0 && (d = dfs(edge[i].v, min(delte, edge[i].w))) > 0) {
edge[i].w -= d; edge[i ^ 1].w += d;
delte -= d;
if(delte == 0) break;
}
}
return flow - delte;
}
int dinic() {
int ans = 0;
while(bfs()) {
for (int i = 1; i <= n; i ++)
cur[i] = pre[i];
ans += dfs(si, INF);
}
return ans;
}
int main(int argc, const char * argv[]) {
while(scanf("%d %d", &n, &m) && (n || m)){
init();
for (int i = 0; i < m; i ++) {
int s, e, w;
scanf("%d %d %d", &s, &e, &w);
a[i] = s; b[i] = e;
addEdge(s, e, w);
addEdge(e, s, w);
}
dinic();
for (int i = 0; i < m; i ++)
if((dis[a[i]] == -1 && dis[b[i]] != -1) || (dis[b[i]] == -1 && dis[a[i]] != -1))
printf("%d %d\n", a[i], b[i]);
printf("\n");
}
return 0;
}