P1345 Telecowmunication

题目链接

这道题是删点
那我们想办法把它变成删边的题,就是删除某条边和删一个点效果一样
把每个点\(i\)拆成\(i+n\)\(i\)\(i\)连入边,\(i+n\)连所有出边。
再把\(i\)\(i+n\)连上一条边。
那么删除\(i\)\(i+n\)间的边和删除\(i\)点效果一样
就变成了求最小割的题

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#include <bits/stdc++.h>
using namespace std;

const int N = 205, M = 2605, inf = 1 << 25;
int n, m, s, t, head[N], cnt = 1, dis[N], now[N];
struct E {
int nxt, v, w;
} e[M];

void add(int u, int v, int w) {
e[++cnt] = {head[u], v, w}; head[u] = cnt;
e[++cnt] = {head[v], u, 0}; head[v] = cnt;
}
bool bfs() {
queue<int> q;
for (int i = 1; i <= 2 * n; i++) dis[i] = inf;
dis[s] = 0; now[s] = head[s];
q.push(s);
while (!q.empty()) {
int u = q.front(); q.pop();
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].v;
if (dis[v] == inf && e[i].w > 0) {
dis[v] = dis[u] + 1;
now[v] = head[v];
q.push(v);
if (v == t) return 1;
}
}
}
return 0;
}
int dfs(int u, int sum) {
if (u == t) return sum;
int k, res = 0;
for (int i = now[u]; i && sum; i = e[i].nxt) {
now[u] = i;
int v = e[i].v;
if (dis[v] == dis[u] + 1 && e[i].w > 0) {
k = dfs(v, min(sum, e[i].w));
if (k == 0) dis[v] = inf;
e[i].w -= k;
e[i ^ 1].w += k;
sum -= k;
res += k;
}
}
return res;
}
int dinic(int s, int t) {
int res = 0;
while (bfs()) res += dfs(s, inf);
return res;
}

int main() {
scanf("%d %d %d %d", &n, &m, &s, &t);
for (int i = 1; i <= n; i++) add(i, i + n, 1);
for (int i = 1; i <= m; i++) {
int u, v;
scanf("%d %d", &u, &v);
add(u + n, v, inf);
add(v + n, u, inf);
}
s += n;
printf("%d\n", dinic(s, t));
return 0;
}


P1345 Telecowmunication
https://widsnoy.top/posts/2879/
作者
widsnoy
发布于
2020年9月16日
许可协议