hdu 1690 构图后Floyd 数据很大

WA了好多次... 这题要用long long 而且INF要设大一点

Sample Input
2 //T
1 2 3 4 1 3 5 7 //L1-L4 C1-C4 距离和花费
4 2 //结点数 询问次数
1 //结点的横坐标
2
3
4
1 4 //起点 终点
4 1
1 2 3 4 1 3 5 7
4 1
1
2
3
10
1 4

Sample Output
Case 1:
The minimum cost between station 1 and station 4 is 3.
The minimum cost between station 4 and station 1 is 3.
Case 2:
Station 1 and station 4 are not attainable.

 # include <iostream>
 # include <cstdio>
 # include <cstring>
 # include <algorithm>
 # include <cmath>
 # include <queue>
 # define LL long long
 using namespace std ;

 const LL INF=0x7f7f7f7f7f7f7f7fLL;
 const int MAXN=;

 LL L[] ;
 LL C[] ;
 LL x[MAXN] ;

 LL dis[MAXN][MAXN];
 int n ;

 void floyed()//节点从1~n编号
 {
     int i,j,k;
     for(k=;k<=n;k++)
        for(i=;i<=n;i++)
          for(j=;j<=n;j++)
              if(dis[i][k]+dis[k][j] < dis[i][j] && dis[i][k] != INF && dis[k][j] != INF)
                  dis[i][j]=dis[i][k]+dis[k][j];

 }

 LL Cost(LL d)
 {
     if (d < )
         d *= - ;
     if (d >  && d<= L[])
         return C[] ;
     if (d > L[] && d<= L[])
         return C[] ;
     if (d > L[] && d<= L[])
         return C[] ;
     if (d > L[] && d<= L[])
         return C[] ;
     return INF ;
 }

 int main ()
 {
    // freopen("in.txt","r",stdin) ;
     int cnt ;
     int T ;
     cin>>T ;
     int Case =  ;
     while (T--)
     {
         Case++ ;
         cout<<"Case "<<Case<<":"<<endl ;
         int i , j  ;
         LL w ;
         for (i =  ; i <=  ; i++)
             cin>>L[i];
         for (i =  ; i <=  ; i++)
             cin>>C[i];

         cin>>n>>cnt ;
         for (i =  ; i <= n ; i++)
             cin>>x[i];
         for (i =  ; i <= n ; i++)
             for (j =  ; j <= n ; j++)
           {
               if(i==j)dis[i][j]=;
               else dis[i][j]=INF;
           }
         for (i =  ; i <= n ; i++)
             for (j = i+ ; j <= n ; j++)
         {
             LL d = x[i] - x[j] ;
             w = Cost(d) ;
             dis[i][j] = w ;
             dis[j][i] = w ;
         }
         floyed() ;
         int u , v ;
         while(cnt--)
         {
             cin>>u>>v ;
             if (dis[u][v] != INF)
               cout<<"The minimum cost between station "<<u<<" and station "<<v<<" is "<<dis[u][v]<<"."<<endl ;
             else
               cout<<"Station "<<u<<" and station "<<v<<" are not attainable."<<endl ;
         }
     }

     return  ;
 }