Cows are such finicky eaters. Each cow has a preference for certain foods and drinks, and she will consume no others.

Farmer John has cooked fabulous meals for his cows, but he forgot to check his menu against their preferences. Although he might not be able to stuff everybody, he wants to give a complete meal of both food and drink to as many cows as possible.

Farmer John has cooked F (1 ≤ F ≤ 100) types of foods and prepared D (1 ≤ D ≤ 100) types of drinks. Each of his N (1 ≤ N ≤ 100) cows has decided whether she is willing to eat a particular food or drink a particular drink. Farmer John must assign a food type and a drink type to each cow to maximize the number of cows who get both.

Each dish or drink can only be consumed by one cow (i.e., once food type 2 is assigned to a cow, no other cow can be assigned food type 2).

 Select Code

#include<cstdio>#include<cstring>#include<iostream>#define EF if(ch==EOF) return x;using namespace std;const int N=405;const int M=1e5+5;const int inf=2e9;struct edge{int v,next,cap;}e[M<<1];int tot=1,head[N];int n,m,ans,f,d,S,T,dis[N],q[N*2];inline int read(){	int x=0,f=1;char ch=getchar();	while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;EF;ch=getchar();}	while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}	return x*f;}void add(int x,int y,int z){	e[++tot].v=y;e[tot].cap=z;e[tot].next=head[x];head[x]=tot;	e[++tot].v=x;e[tot].cap=0;e[tot].next=head[y];head[y]=tot;}bool bfs(){	memset(dis,-1,sizeof dis);	int h=0,t=1;q[t]=S;dis[S]=0;	while(h!=t){		int x=q[++h];		for(int i=head[x];i;i=e[i].next){			if(e[i].cap&&dis[e[i].v]==-1){				dis[e[i].v]=dis[x]+1;				if(e[i].v==T) return 1;				q[++t]=e[i].v;			}		} 	}	return 0;}int dfs(int x,int f){	if(x==T) return f;	int used=0,t;	for(int i=head[x];i;i=e[i].next){		if(e[i].cap&&dis[e[i].v]==dis[x]+1){			t=dfs(e[i].v,min(e[i].cap,f));			e[i].cap-=t;e[i^1].cap+=t;			used+=t;f-=t;			if(!f) return used;		}	}	if(!used) dis[x]=-1;	return used;}void dinic(){	ans=0;	while(bfs()) ans+=dfs(S,inf);}void work(){	n=read();f=read();d=read();S=0;T=f+n+n+d+1;	for(int i=1,fi,di,x;i<=n;i++){		fi=read();		di=read();		while(fi--){			x=read();			add(x,f+i,1);		}		while(di--){			x=read();			add(f+n+i,f+n+n+x,1);		}	}	for(int i=1;i<=n;i++) add(f+i,f+n+i,1);	for(int i=1;i<=f;i++) add(S,i,1);	for(int i=1;i<=d;i++) add(f+n+n+i,T,1);	dinic();	printf("%d\n",ans);}int main(){	work();	return 0;}