题意:给一个n点m条边的无向图,现在要删掉一些边使得点1、2不连通,每条边有一个花费,问如何删边
思路:显而易见求最小割,也就是最大流,主要是如何输出路径
引用自https://blog.csdn.net/ac_lion/article/details/8620676
最小割,就是在所有割中,容量之和最小的割。最小割的值就是最大流的值,因为很容易想到,从源点s到汇点t的最大流必然会经过割边,那么就有最大流f<=c(割边的值),那么也就是说,当c==f的时候,就是c为小割,即最大流==最小割
在跑完最大流后的残余网络上,记录源点和汇点可达的点,然后遍历所有的边(设边的两端点为 u 和 v),如果 u 可达源点而 v 可达汇点(或与之相反,v 可达源点而 u 可达汇点),则说明该边必须割。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int N = 55;
const int M = 2005;
int flag[N];
int head[N], dis[N], tot, n, m, maxflow;
struct Edge {
int from, to, next, cap, flow;
}edge[M];
void init() {
tot = 2;
memset(head, -1, sizeof(head));
memset(flag, 0, sizeof(flag));
}
//rw为反向流量,单向边addedge(u, v, w)
//双向边时正反向流量相同可直接addedge(u, v, w, w)
//双向边若加两条边(憨批操作)addedge(u, v, w), addedge(v, u, w), 输出路径时要注意每次+4,即for(int i = 0; i < tot; i += 4)
void addedge(int u, int v, int w, int rw = 0) {
edge[tot].from = u;
edge[tot].to = v;
edge[tot].cap = w;
edge[tot].flow = 0;
edge[tot].next = head[u];
head[u] = tot++;
edge[tot].from = v;
edge[tot].to = u;
edge[tot].cap = rw;
edge[tot].flow = 0;
edge[tot].next = head[v];
head[v] = tot++;
}
int Q[N];
int dep[N], cur[N], sta[N]; ///数组cur记录点u之前循环到了哪一条边
bool bfs(int s, int t, int n) {
int fron = 0, tail = 0;
memset(dep, -1, sizeof(dep[0]) * (n + 1));
dep[s] = 0;
Q[tail++] = s;
while(fron < tail) {
int u = Q[fron++];
for(int i = head[u]; i != -1; i = edge[i].next) {
int v = edge[i].to;
if(edge[i].cap > edge[i].flow && dep[v] == -1) {
dep[v] = dep[u] + 1;
if(v == t) return true;
Q[tail++] = v;
}
}
}
return false;
}
void dinic(int s, int t, int n) { //源点 汇点 点数
maxflow = 0;
while(bfs(s, t, n)) {
for(int i = 0; i <= n; ++i) cur[i] = head[i];
int u = s, tail = 0;
while(cur[s] != -1) {
if(u == t) {
int tp = inf;
for(int i = tail - 1; i >= 0; --i)
tp = min(tp, edge[sta[i]].cap - edge[sta[i]].flow);
maxflow += tp;
for(int i = tail - 1; i >= 0; --i) {
edge[sta[i]].flow += tp;
edge[sta[i] ^ 1].flow -= tp;
if(edge[sta[i]].cap - edge[sta[i]].flow == 0)
tail = i;
}
u = edge[sta[tail] ^ 1].to;
}
else if(cur[u] != -1 && edge[cur[u]].cap > edge[cur[u]].flow && dep[u] + 1 == dep[edge[cur[u]].to]) {
sta[tail++] = cur[u];
u = edge[cur[u]].to;
}
else {
while(u != s && cur[u] == -1)
u = edge[sta[--tail] ^ 1].to;
cur[u] = edge[cur[u]].next;
}
}
}
}
void dfs(int u, int op) {
flag[u] = op;
for(int i = head[u]; ~i; i = edge[i].next) {
if(edge[i].flow == edge[i].cap) continue;
int v = edge[i].to;
if(!flag[v]) dfs(v, op);
}
}
int main(){
// freopen("in.txt", "r", stdin);
while(~scanf("%d%d", &n, &m) && n + m) {
init();
int u, v, w;
while(m--) {
scanf("%d%d%d", &u, &v, &w);
addedge(u, v, w, w);
}
dinic(1, 2, n);
dfs(1, 1);
dfs(2, 2);
for(int i = 0; i < tot; i += 2) {
int u = edge[i].from, v = edge[i].to;
if(flag[u] + flag[v] == 3)
printf("%d %d\n", u, v);
}
printf("\n");
}
return 0;
}