传送门：点击打开链接
题意：给一个DAG模型，要求最少的不相交的路径覆盖所有的点，即最小路径覆盖。
思路：最小路径覆盖通常把点拆成入点和出点，然后得到最大匹配数，答案就等于原点个数-最大匹配数
用二分图求最小路径覆盖的前提是，只能是DAG模型，不能有环！
#include<map>
#include<set>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<cstdio>
#include<cctype>
#include<bitset>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define fuck(x) cout<<"["<<x<<"]"
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)
using namespace std;
typedef long long LL;const int MX = 2e3 + 5;
const int MS = 2e5 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;int Head[MX], erear;
struct Edge {int v, nxt;
} E[MS];
void edge_init() {erear = 0;memset(Head, -1, sizeof(Head));
}
void edge_add(int u, int v) {E[erear].v = v;E[erear].nxt = Head[u];Head[u] = erear++;
}int match[MX];
bool vis[MX];
bool DFS(int u) {for(int i = Head[u]; ~i; i = E[i].nxt) {int v = E[i].v;if(!vis[v]) {vis[v] = 1;if(match[v] == -1 || DFS(match[v])) {match[v] = u;return 1;}}}return 0;
}
int BM(int n) {int res = 0;memset(match, -1, sizeof(match));for(int u = 1; u <= n; u++) {memset(vis, 0, sizeof(vis));if(DFS(u)) res++;}return res;
}int A[MX];
int main() {int T; //FIN;scanf("%d", &T);while(T--) {int n, m;edge_init();scanf("%d%d", &n ,&m);for(int i = 1; i <= m; i++) {int u, v;scanf("%d%d", &u, &v);edge_add(u, n + v);}printf("%d\n", n - BM(n));}return 0;
}