A simple brute force problem.

  1 #include<iostream>
  2 #include<cstring>
  3 #include<algorithm>
  4 #include<cstdio>
  5 #include<queue>
  6 #define INF 100000000
  7 using namespace std;
  8 const int N=205;
  9 const int M=1000000;
 10 struct Node{
 11    int v;
 12    int f;
 13    int next;
 14 }edge[M];
 15 int n,m,u,v,cnt,sx,ex;
 16 int head[N],pre[N],visit[N];
 17 int val1[N],val2[N];
 18 void init(){
 19     cnt=0;
 20     //memset(edge,0,sizeof(edge));
 21     memset(head,-1,sizeof(head));
 22 }
 23 void add(int u,int v,int w){
 24     edge[cnt].v=v;
 25     edge[cnt].f=w;
 26     edge[cnt].next=head[u];
 27     head[u]=cnt++;
 28     edge[cnt].f=0;
 29     edge[cnt].v=u;
 30     edge[cnt].next=head[v];
 31     head[v]=cnt++;
 32 }
 33 bool BFS(){
 34   memset(pre,0,sizeof(pre));
 35   pre[sx]=1;
 36   queue<int>Q;
 37   Q.push(sx);
 38  while(!Q.empty()){
 39      int d=Q.front();
 40      Q.pop();
 41      for(int i=head[d];i!=-1;i=edge[i].next    ){
 42         if(edge[i].f&&!pre[edge[i].v]){
 43             pre[edge[i].v]=pre[d]+1;
 44             Q.push(edge[i].v);
 45         }
 46      }
 47   }
 48  return pre[ex]>0;
 49 }
 50 int dinic(int flow,int ps){
 51     int f=flow;
 52      if(ps==ex)return f;
 53      for(int i=head[ps];i!=-1;i=edge[i].next){
 54         if(edge[i].f&&pre[edge[i].v]==pre[ps]+1){
 55             int a=edge[i].f;
 56             int t=dinic(min(a,flow),edge[i].v);
 57               edge[i].f-=t;
 58               edge[i^1].f+=t;
 59             flow-=t;
 60              if(flow<=0)break;
 61         }
 62 
 63      }
 64       if(f-flow<=0)pre[ps]=-1;
 65       return f-flow;
 66 }
 67 int solve(){
 68     int sum=0;
 69     while(BFS())
 70         sum+=dinic(INF,sx);
 71     return sum;
 72 }
 73 int main() {
 74     int T,k,temp,sum,tt=1;
 75     scanf("%d",&T);
 76     while(T--) {
 77          scanf("%d%d",&n,&m);
 78          sum=0;
 79          init();
 80         for(int i=1; i<=n; i++) {
 81             scanf("%d",&val1[i]);
 82             sum+=val1[i];
 83         }
 84        for(int j=1;j<=m;j++)
 85           scanf("%d",&val2[j]);
 86        for(int i=1;i<=n;i++){
 87           scanf("%d",&k);
 88           for(int j=1;j<=k;j++){
 89             scanf("%d",&temp);
 90             add(2*m+i,temp+1,INF);
 91           }
 92           add(0,2*m+i,val1[i]);
 93        }
 94        for(int i=1;i<=m;i++){
 95           for(int j=1;j<=m;j++){
 96               scanf("%d",&temp);
 97               if(temp==1){
 98                 add(i,j,INF);
 99               }
100           }
101        }
102        for(int i=1;i<=m;i++){
103           add(i,i+m,val2[i]);
104           add(i+m,2*m+n+1,INF);
105        }
106        sx=0;ex=2*m+n+1;
107        printf("Case #%d: %d\n",tt++,sum-solve());
108     }
109     return 0;
110 }