POJ1422

题目大意：

一个有向无环图，求最小路径覆盖

板子题。。。

把每个点拆成\(x\)和\(x^{'}\)

\((u,v)\)有一条边则\(u\)向\(v^{'}\)连一条边，然后跑最大匹配，最小路径覆盖=节点数-最大匹配

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
namespace red{
#define eps (1e-8)
	inline int read()
	{
		int x=0;char ch,f=1;
		for(ch=getchar();(ch<'0'||ch>'9')&&ch!='-';ch=getchar());
		if(ch=='-') f=0,ch=getchar();
		while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
		return f?x:-x;
	}
	const int N=410;
	int haku;
	int n,m,ret,tim;
	int f[N],vis[N];
	int head[N],cnt;
	struct point
	{
		int nxt,to;
		point(){}
		point(const int &nxt,const int &to):nxt(nxt),to(to){}
	}a[N*N];
	inline void link(int x,int y)
	{
		a[++cnt]=(point){head[x],y};head[x]=cnt;
		a[++cnt]=(point){head[y],x};head[y]=cnt;
	}
	inline bool find(int x)
	{
		for(int i=head[x];i;i=a[i].nxt)
		{
			int t=a[i].to;
			if(vis[t]==tim) continue;
			vis[t]=tim;
			if(!f[t]||find(f[t]))
			{
				f[t]=x;
				return 1;
			}
		}
		return 0;
	}
	inline void main()
	{
		haku=read();
		while(haku--)
		{
			memset(head,0,sizeof(head));
			memset(f,0,sizeof(f));
			cnt=ret=0;
			n=read(),m=read();
			for(int x,y,i=1;i<=m;++i)
			{
				x=read(),y=read();
				link(x,y+n);
			}
			for(int i=1;i<=n;++i)
			{
				++tim;
				ret+=find(i);
			}
			printf("%d\n",n-ret);
		}
	}
}
signed main()
{
	red::main();
return 0;
}