Alice's Chance

  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=805;
  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];
 17 int val1[N],val2[N];//根据题目要求申请
 18 void init(){
 19     cnt=0;
 20     memset(head,-1,sizeof(head));
 21 }
 22 void add(int u,int v,int w){
 23     edge[cnt].v=v;
 24     edge[cnt].f=w;
 25     edge[cnt].next=head[u];
 26     head[u]=cnt++;
 27     edge[cnt].f=0;
 28     edge[cnt].v=u;
 29     edge[cnt].next=head[v];
 30     head[v]=cnt++;
 31 }
 32 bool BFS(){
 33   memset(pre,0,sizeof(pre));
 34   pre[sx]=1;
 35   queue<int>Q;
 36   Q.push(sx);
 37  while(!Q.empty()){
 38      int d=Q.front();
 39      Q.pop();
 40      for(int i=head[d];i!=-1;i=edge[i].next    ){
 41         if(edge[i].f&&!pre[edge[i].v]){
 42             pre[edge[i].v]=pre[d]+1;
 43             Q.push(edge[i].v);
 44         }
 45      }
 46   }
 47  return pre[ex]>0;
 48 }
 49 int dinic(int flow,int ps){
 50     int f=flow;
 51      if(ps==ex)return f;
 52      for(int i=head[ps];i!=-1;i=edge[i].next){
 53         if(edge[i].f&&pre[edge[i].v]==pre[ps]+1){
 54             int a=edge[i].f;
 55             int t=dinic(min(a,flow),edge[i].v);
 56               edge[i].f-=t;
 57               edge[i^1].f+=t;
 58             flow-=t;
 59              if(flow<=0)break;
 60         }
 61 
 62      }
 63       if(f-flow<=0)pre[ps]=-1;
 64       return f-flow;
 65 }
 66 int solve(){
 67     int sum=0;
 68     while(BFS())
 69         sum+=dinic(INF,sx);
 70     return sum;
 71 }
 72 int a[8];
 73 int main() {
 74     int T,k,temp,sum,tt=0,t1,t2;
 75     scanf("%d",&T);
 76     while(T--) {
 77          scanf("%d",&n);tt=0;
 78          init();sum=0;sx=0,ex=380;
 79         for(int i=1;i<=n; i++) {
 80             for(int j=1;j<=7;j++){
 81                scanf("%d",&a[j]);
 82             }
 83             scanf("%d%d",&t1,&t2);
 84             sum+=t1;
 85           for(int j=1;j<=7;j++){
 86             if(a[j]==1){
 87                 for(int k=0;k<t2;k++)
 88                  add(i+350,k*7+j,1);
 89             }
 90           }
 91           add(0,i+350,t1);
 92         }
 93         for(int i=1;i<=350;i++)
 94             add(i,380,1);
 95         if(sum==solve()){
 96             printf("Yes\n");
 97         }
 98         else printf("No\n");
 99 
100 
101     }
102     return 0;
103 }