/*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 0x3fffffffstruct 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\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;}
