poj 1698 Alice's Chance 最大流

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_JUDGE

    freopen("sum.in","r",stdin);

    //freopen("sum.out","w",stdout);

#endif



    int 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;

}