Luogu P7368 [USACO05NOV]Asteroids G

题目描述

https://www.luogu.com.cn/problem/P7368

Solution

我们考虑将行和列分别看成二分图的左部和右部

如果存在点 $(x,y)$，则意味着行 $x$ 和列 $y$ 至少要消除一个

我们发现这个的东西就是二分图最小点覆盖，所以我们求一个二分图最大匹配就行了

#include <iostream>
#include <cstdio>
#include <queue>
#define maxn 1010
#define INF 1000000000
using namespace std;

int n, m, s, t;

struct Edge {
    int to, next, w;
} e[1000000]; 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]; 
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 (u == t || !_w) 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 dinic() {
    int mf = 0;
    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 = 0; t = 2 * n + 1; 
    for (int i = 1; i <= n; ++i) { 
        Add_edge(s, i, 1); 
        Add_edge(i + n, t, 1);
    }
    for (int i = 1; i <= m; ++i) {
        int x, y; cin >> x >> y;
        Add_edge(x, y + n, 1); 
    } cout << dinic() << "\n"; 
    return 0;
}