POJ 1698 Alice's Chance

/* 一个演员很想参加film，但是一天只能参加一项，且每项必须在规定的几个星期内参加规定的天数，求演员是否能全部完成所有的film档期。 可以考虑用网络流做，加源汇点，如果film_i 和week_i_day_j天有联系，那么就连一条INF边，源点到film_i连一条day_i边，各个星期的天数到汇点连一条1边，求最大流是否和总天数相等。最大流用isap就可以了，非常快。 */

 

#include  < iostream >
#include  < cstdio >
#include  < algorithm >
#include  < memory.h >
#include  < queue >
using   namespace  std;

#define  INF 0x7ffffff
#define  MAXN 500
#define  MAXE 20000

struct  Edge{
     int  next,pair;
     int  v,cap,flow;
}edge[MAXE];
int  net[MAXN];
int  nv,ne,s,t,index,max_week,sum;
void  init()
{
    index  =   0 ;
    memset(net, - 1 , sizeof (net));
    s  =   0 ;  //  start position
    max_week  =   - 1 ;
    sum  =   0 ;
}
void  add_edge( const   int &  u, const   int &  v, const   int &  val)
{
    edge[index].next  =  net[u];
    net[u]  =  index;
    edge[index].v  =  v;
    edge[index].cap  =  val;
    edge[index].flow  =   0 ;
    edge[index].pair  =  index  +   1 ;
     ++ index;
    edge[index].next  =  net[v];
    net[v]  =  index;
    edge[index].v  =  u;
    edge[index].cap  =   0 ;
    edge[index].flow  =   0 ;
    edge[index].pair  =  index  -   1 ;
     ++ index;
}
int  ISAP()
{
     int  numb[MAXN],dist[MAXN],curedge[MAXN],pre[MAXN];
     int  cur_flow,max_flow,u,tmp,neck,i;
    memset(dist, 0 , sizeof (dist));
    memset(numb, 0 , sizeof (numb));
     for (i  =   1  ; i  <=  nv ;  ++ i)
        curedge[i]  =  net[i];
    numb[nv]  =  nv;
    max_flow  =   0 ;
    u  =  s;
     while (dist[s]  <  nv){
         if (u  ==  t){
            cur_flow  =  INF + 1 ;
             for (i  =  s; i  !=  t;i  =  edge[curedge[i]].v) 
                 if (cur_flow  >  edge[curedge[i]].cap){
                    neck  =  i;
                    cur_flow  =  edge[curedge[i]].cap;
                }
             for (i  =  s; i  !=  t; i  =  edge[curedge[i]].v)
            {
                tmp  =  curedge[i];
                edge[tmp].cap  -=  cur_flow;
                edge[tmp].flow  +=  cur_flow;
                tmp  =  edge[tmp].pair;
                edge[tmp].cap  +=  cur_flow;
                edge[tmp].flow  -=  cur_flow;
            }
            max_flow  +=  cur_flow;
            u  =  neck;
        }
         for (i  =  curedge[u]; i  !=   - 1 ; i  =  edge[i].next)
             if (edge[i].cap  >   0   &&  dist[u]  ==  dist[edge[i].v] + 1 )
                 break ;
         if (i  !=   - 1 ){
            curedge[u]  =  i;
            pre[edge[i].v]  =  u;
            u  =  edge[i].v;
        } else {
             if ( 0   ==   -- numb[dist[u]])  break ;
            curedge[u]  =  net[u];
             for (tmp  =  nv,i  =  net[u]; i  !=   - 1 ; i  =  edge[i].next)
                 if (edge[i].cap  >   0 )
                    tmp  =  tmp < dist[edge[i].v] ? tmp:dist[edge[i].v];
            dist[u]  =  tmp  +   1 ;
             ++ numb[dist[u]];
             if (u  !=  s) u  =  pre[u];
        }
    }
     return  max_flow;
}

int  main()
{
     int  i,j,k,h,time,ans,n,work[ 7 ],day,week;
    scanf( " %d " , & time);
     while (time -- )
    {
        init();
        scanf( " %d " , & n);
         for (i  =   1 ; i  <=  n;  ++ i)
        {
             for (j  =   0 ;j  <   7 ;  ++ j)
                scanf( " %d " , & work[j]);
            scanf( " %d%d " , & day, & week);
            add_edge(s,i,day);
             for (j  =   0 ;j  <  week;  ++ j)
                 for (k  =   0 ;k  <   7 ;  ++ k)
                     if (work[k])  //  can do 
                        add_edge(i,n + j * 7 + k + 1 ,INF);
            max_week  =  max_week > week ? max_week:week;
            sum  +=  day;
        }
        t  =  n + max_week * 7 + 1 ;
        nv  =  t  +   1 ;
         for (i  =   0 ;i  <  max_week;  ++ i )
             for (j  =   1 ;j  <=   7 ;  ++ j)
                add_edge(n + i * 7 + j,t, 1 );
        ans  =  ISAP();
         if (sum  ==  ans)
            printf( " Yes\n " );
         else
            printf( " No\n " );
    }
     return   0 ;
}