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

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
#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;
}