匈牙利算法板子

复杂度$O(VE)$, 处理完后, $f$数组前$n$位一定为$0$, 后$m$位存储对应匹配.

const int N = 2e3+10;
int n, m, e, clk, f[N], vis[N];
vector<int> g[N];
int dfs(int x) {
	for (int y:g[x]) if (vis[y]!=clk) {
		vis[y] = clk;
		if (!f[y]||dfs(f[y])) return f[y]=x;
	}
	return 0;
}
int main() {
	scanf("%d%d%d", &n, &m, &e);
	REP(i,1,e) { 
		int u, v;
		scanf("%d%d", &u, &v);
		if (u>n||v>m) continue;
		g[u].pb(v+n);
		g[v+n].pb(u);
	}
	int ans = 0;
	REP(i,1,n) {
		++clk;
		if (dfs(i)) ++ans;
	}
	printf("%d\n", ans);
}

 

求有向无环图最小路径覆盖.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif

int n, m, clk, f[N], vis[N];
int deg[N], nxt[N];
vector<int> g[N];
int dfs(int x) {
	for (int y:g[x]) if (vis[y]!=clk) { 
		vis[y]=clk;
		if (!f[y]||dfs(f[y])) return f[y]=x;
	}
	return 0;
}
int main() {
	scanf("%d%d", &n, &m);
	REP(i,1,m) {
		int u, v;
		scanf("%d%d", &u, &v);
		g[u].pb(v+n);
		g[v+n].pb(u);
	}
	int ans = 0;
	REP(i,1,n) {
		++clk;
		if (dfs(i)) ++ans;
	}
	REP(i,n+1,2*n) if (f[i]) nxt[f[i]]=i-n,deg[i-n]=1;
	REP(i,1,n) if (!deg[i]) {
		int u = i;
		do printf("%d ",u),u=nxt[u]; while (u);
		hr;
	}
	printf("%d\n", n-ans);
}