Luogu P1345 [USACO5.4]奶牛的电信Telecowmunication

题目描述

https://www.luogu.com.cn/problem/P1345

简要题意：给定一张 $n$ 个点 $m$ 条边的无向图和起点 $s$ 以及终点 $t$，求最少删掉几个点使得 $s$ 和 $t$ 不在连通

$n\le 100,m\le 600$

Solution

建图：

$u$ 连 $u’$，容量为 $1$

原图中的双向边，$u’$ 连 $v$，$v’$ 连 $u$，容量均为 $\infty$

#include <iostream>
#include <cstdio>
#include <queue>
#define maxn 210
#define maxm 610
#define INF 1000000000
using namespace std;

int n, m;

struct Edge {
    int to, next, w;
} e[maxm * 6]; int c1, head[maxn];
inline void add_edge(int u, int v, int w) {
    e[c1].to = v; e[c1].w = w;
    e[c1].next = head[u]; head[u] = c1++;
}

inline void Add_edge(int u, int v, int w) {
    add_edge(u, v, w); add_edge(v, u, 0);
}

int dep[maxn], s, t; 
bool bfs() {
    fill(dep, dep + maxn, 0); dep[s] = 1;
    queue<int> Q; Q.push(s);
    while (!Q.empty()) {
        int u = Q.front(); Q.pop(); 
        for (int i = head[u]; ~i; i = e[i].next) {
            int v = e[i].to, w = e[i].w;
            if (w > 0 && !dep[v]) {
                dep[v] = dep[u] + 1;
                Q.push(v); if (v == t) return 1; 
            }
        }
    }
    return 0; 
}

int dfs(int u, int _w) {
    if (!_w || u == t) return _w;
    int s = 0;
    for (int i = head[u]; ~i; i = e[i].next) {
        int v = e[i].to, w = e[i].w;
        if (w > 0 && dep[v] == dep[u] + 1) {
            int di = dfs(v, min(_w - s, w));
            e[i].w -= di; e[i ^ 1].w += di;
            s += di; if (s == _w) break; 
        }
    }
    if (s < _w) dep[u] = 0; return s; 
}

int mf;
int dinic() {
    while (bfs()) mf += dfs(s, INF);
    return mf;
}

int main() { fill(head, head + maxn, -1); 
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    cin >> n >> m >> s >> t; s += n;
    for (int i = 1; i <= n; ++i) Add_edge(i, i + n, 1);
    for (int i = 1; i <= m; ++i) {
        int x, y; cin >> x >> y;
        Add_edge(x + n, y, INF); Add_edge(y + n, x, INF); 
    } cout << dinic() << "\n";
    return 0;
}