题目：给出n部电影的可以在周几拍摄、总天数、期限，问能不能把n部电影接下来。
分析：
对于每部电影连上源点，流量为总天数。
对于每一天建立一个点，连上汇点，流量为为1。
对于每部电影，如果可以在该天拍摄，则连上一条流量为1的边。
跑一次最大流。。。
#include <set>
#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>using namespace std;typedef long long ll;
typedef unsigned long long ull;#define debug puts("here")
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define foreach(i,vec) for(unsigned i=0;i<vec.size();i++)
#define pb push_back
#define RD(n) scanf("%d",&n)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define RD4(x,y,z,w) scanf("%d%d%d%d",&x,&y,&z,&w)
#define All(vec) vec.begin(),vec.end()
#define MP make_pair
#define PII pair<int,int>
#define PQ priority_queue
#define cmax(x,y) x = max(x,y)
#define cmin(x,y) x = min(x,y)
#define Clear(x) memset(x,0,sizeof(x))
/*#pragma comment(linker, "/STACK:1024000000,1024000000")int size = 256 << 20; // 256MB
char *p = (char*)malloc(size) + size;
__asm__("movl %0, %%esp\n" :: "r"(p) );*//******** program ********************/const int MAXN = 1005;
const int MAXM = 100005;
const int INF = 1e9;int po[MAXN],tol;
int gap[MAXN],dis[MAXN],arc[MAXN],pre[MAXN],cur[MAXN];
int n,m,vs,vt;struct Edge{int y,f,next;
}edge[MAXM];void Add(int x,int y,int f){edge[++tol].y = y;edge[tol].f = f;edge[tol].next = po[x];po[x] = tol;
}
void add(int x,int y,int f){Add(x,y,f);Add(y,x,0);
}int sap(){memset(dis,0,sizeof(dis));memset(gap,0,sizeof(gap));gap[0] = vt;rep1(i,vt)arc[i] = po[i];int ans = 0;int aug = INF;int x = vs;while(dis[vs]<vt){bool ok = false;cur[x] = aug;for(int i=arc[x];i;i=edge[i].next){int y = edge[i].y;if(edge[i].f>0&&dis[y]+1==dis[x]){ok = true;pre[y] = arc[x] = i;aug = min(aug,edge[i].f);x = y;if(x==vt){ans += aug;while(x!=vs){edge[pre[x]].f -= aug;edge[pre[x]^1].f += aug;x = edge[pre[x]^1].y;}aug = INF;}break;}}if(ok)continue;int MIN = vt-1;for(int i=po[x];i;i=edge[i].next)if(edge[i].f>0&&dis[edge[i].y]<MIN){MIN = dis[edge[i].y];arc[x] = i;}if(--gap[dis[x]]==0)break;dis[x] = ++ MIN;++ gap[dis[x]];if(x!=vs){x = edge[pre[x]^1].y;aug = cur[x];}}return ans;
}int f[10],w,d;int main(){#ifndef ONLINE_JUDGEfreopen("sum.in","r",stdin);//freopen("sum.out","w",stdout);
#endifint ncase;RD(ncase);while(ncase--){Clear(po);tol = 1;vs = MAXN-3;vt = vs+1;int ans = 0;int t = 0;RD(n);rep1(i,n){rep(j,7)RD(f[j]);RD2(d,w);ans += d;add(vs,i,d);w *= 7;rep(j,w)if(f[j%7])add(i,j+n+1,1);cmax(t,w);}rep1(i,t)add(i+n,vt,1);ans -= sap();puts(ans==0?"Yes":"No");}return 0;
}
