[USACO2007OPENG] Dining G

题目描述

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).

输入格式

Line 1: Three space-separated integers: N, F, and D

Lines 2..N+1: Each line i starts with a two integers Fi and Di, the number of dishes that cow i likes and the number of drinks that cow i likes. The next Fi integers denote the dishes that cow i will eat, and the Di integers following that denote the drinks that cow i will drink.

输出格式

Line 1: A single integer that is the maximum number of cows that can be fed both food and drink that conform to their wishes

样例 #1

样例输入 #1

4 3 3
2 2 1 2 3 1
2 2 2 3 1 2
2 2 1 3 1 2
2 1 1 3 3

样例输出 #1

3

提示

One way to satisfy three cows is:

Cow 1: no meal

Cow 2: Food #2, Drink #2

Cow 3: Food #1, Drink #1

Cow 4: Food #3, Drink #3

The pigeon-hole principle tells us we can do no better since there are only three kinds of food or drink. Other test data sets are more challenging, of course.

看到这么小的数据，还有这种选来选去的东西，考虑网络流。

在建图中把牛放在中间，前面放食物，后面放饮料。

由源点向每个食物放一条流量为1的边，每个饮料向汇点放一条流量为1的边。然后把牛拆成两份，中间流量为1.牛和喜欢的饮料食物连边即可。跑一次最大流，每个牛，饮料，食物都会选一次。刚好就可以跑出答案。

#include<bits/stdc++.h>
using namespace std;
int idx=1,hd[200005],k,v[200005],q[200005],thd[200005],kt,l,r,n,f,d,x,y,z,ans,p,qq,s,t;
struct edge{
	int v,nxt,f;
}e[2000005];
void add_edge(int u,int v)
{
	e[++idx]=(edge){v,hd[u],1};
	hd[u]=idx;
	e[++idx]=(edge){u,hd[v],0};
	hd[v]=idx;
}
int bfs()
{
	memcpy(hd,thd,sizeof(hd));
	memset(v,0,sizeof(v));
	q[l=r=1]=0,v[0]=1;
	while(l<=r)
	{
		for(int i=hd[q[l]];i;i=e[i].nxt)
		{
			if(e[i].f&&!v[e[i].v])
			{
				v[e[i].v]=v[q[l]]+1;
				q[++r]=e[i].v;
			}
		}
		++l;
	}
	return v[t];
}
int dfs(int x,int flow)
{
	if(x==t)
		return flow;
	for(int&i=hd[x];i;i=e[i].nxt)
	{
		if(v[x]+1==v[e[i].v]&&min(flow,e[i].f))
		{
			if(kt=dfs(e[i].v,min(flow,e[i].f)))
			{
				e[i^1].f+=kt;
				e[i].f-=kt;
				return kt;
			}
		}
	}
	return 0;
}
int main()
{
	scanf("%d%d%d",&n,&f,&d);
	s=0,t=1+f+n+d+1;
	for(int i=1;i<=f;i++){
        add_edge(s,1+i);
    }
    for(int i=1;i<=d;i++){
        add_edge(1+f+n+i,t);
    }
    for(int i=1;i<=n;i++){
        add_edge(1+f+i,1+f+n+d+1+i);
    }
    for(int i=1;i<=n;i++){
        scanf("%d%d",&x,&y);
        for(int j=1;j<=x;j++)
		{
            scanf("%d",&z);
            add_edge(1+z,1+f+i);
        }
        for(int j=1;j<=y;j++)
		{
            scanf("%d",&z);
            add_edge(1+f+n+d+1+i,1+f+n+z);
        }
    }
	memcpy(thd,hd,sizeof(hd));
	while(bfs())
		while(k=dfs(0,2147483647))
			ans+=k;
	printf("%d",ans);
	return 0;
}