poj 3281 最大流

/*
思路：start=0为起点，1~F中每个点为每份food的编号，F+1~F+n中为每头牛的编号，另外F+n+1~F+n+n中也为每头牛的编号，F+n+n+1~F+n+n+D中每个点为每份drink的编号，F+n+n+D+1为终点;
      从start到每份food连边，权值为1,；
      输入每头牛可以吃的food时，从这头牛(F+1~F+n)到能吃的food(1~F)连边，权值为1；
	  从当前牛(F+1~F+n)到当前牛(F+n+1~F+n+n)连边,权值为1;(每头牛只需一份食物)
	  输入每头牛可以喝的drink时，从这头牛(F+n+1~F+n+n)到能喝的drink(F+n+n+1~F+n+n+D)连边，权值为1；
	  从每份drink到终点(end=F+n+n+D+1)连边，权值为1；
	 最后用一般的方法求最大流即可;
*/
#include "stdio.h"   //最大流, poj3281
#include "string.h"
#include "queue"
using namespace std;

#define N 450
#define INF 0x3fffffff

struct node {
	int u,v,w;
	int next;
}edge[4*N*N];

int start,end;
int n,idx;
int dis[N],head[N],route[N];

int EK();
int BFS();
void init();
void adde(int u,int v,int w);
void addedge(int u,int v,int w);
int MIN(int x,int y) { return x<y?x:y;	}

int main()
{
	int F,D;
	int i,j,k,a1,a2;
	while(scanf("%d%d%d",&n,&F,&D)!=-1)
	{
		init();
		start = 0;     //0->起点
		end = 2*n+F+D+1;
		for(i=1;i<=F;i++)  //1~F每个点代表每份food;
			adde(start,i,1);
		for(i=1;i<=n;i++)   //F+1~F+n 每个点代表每头牛; F+n+1~F+n+n 每个点也代表每头牛;
		{
			scanf("%d%d",&a1,&a2);
			for(j=1;j<=a1;j++)
			{
				scanf("%d",&k);  
				adde(k,i+F,1);	//food到每头牛连边
			}
			adde(i+F,i+F+n,1);  //(牛到牛连边)每头牛只要一份食物就够了！
			for(j=1;j<=a2;j++)
			{
				scanf("%d",&k);
				adde(i+F+n,2*n+F+k,1); //牛到drink连边
			}
		}
		for(i=1;i<=D;i++)
			adde(2*n+F+i,end,1);   //drink到终点连边
		int ans = EK();   //最后求最大流
		printf("%d
",ans);
	}
	return 0;
}

void init()
{
	idx = 0;
	memset(head,-1,sizeof(head));
}

void adde(int u,int v,int w)
{
	addedge(u,v,w);
	addedge(v,u,0);
}

void addedge(int u,int v,int w)
{
	edge[idx].u = u;
	edge[idx].v = v;
	edge[idx].w = w;
	edge[idx].next = head[u];
	head[u] = idx;
	idx++;
}

int BFS()
{
	int i;
	int x,y;
	memset(route,-1,sizeof(route));
	dis[0] = INF;
	route[0] = 0;
	queue<int> q;
	q.push(start);
	while(!q.empty())
	{
		x = q.front();
		q.pop();
		for(i=head[x];i!=-1;i=edge[i].next)
		{
			y = edge[i].v;
			if(route[y]==-1 && edge[i].w)
			{
				route[y] = i;
				dis[y] = MIN(dis[x],edge[i].w);
				q.push(y);
			}
		}
	}
	route[0] = -1;
	if(route[end]==-1) return 0;
	return dis[end];
}

int EK()
{
	int x,y;
	int ans=0,kejia;
	while(kejia = BFS())
	{
		ans += kejia;
		y = route[end];
		while(y!=-1)
		{
			x = y^1;
			edge[y].w --;
			edge[x].w ++;
			y = route[edge[y].u];
		}
	}
	return ans;
}