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
,head
,route
;

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\n",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;
}