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=210;

LL L[10] ;
LL C[10] ;
LL x[MAXN] ;

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


void floyed()//节点从1~n编号
{
    int i,j,k;
    for(k=1;k<=n;k++)
       for(i=1;i<=n;i++)
         for(j=1;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 < 0)
        d *= -1 ;
    if (d > 0 && d<= L[1])
        return C[1] ;
    if (d > L[1] && d<= L[2])
        return C[2] ;
    if (d > L[2] && d<= L[3])
        return C[3] ;
    if (d > L[3] && d<= L[4])
        return C[4] ;
    return INF ;
}

int main ()
{
   // freopen("in.txt","r",stdin) ;
    int cnt ;
    int T ;
    cin>>T ;
    int Case = 0 ;
    while (T--)
    {
        Case++ ;
        cout<<"Case "<<Case<<":"<<endl ;
        int i , j  ;
        LL w ;
        for (i = 1 ; i <= 4 ; i++)
            cin>>L[i];
        for (i = 1 ; i <= 4 ; i++)
            cin>>C[i];

        cin>>n>>cnt ;
        for (i = 1 ; i <= n ; i++)
            cin>>x[i];
        for (i = 1 ; i <= n ; i++)
            for (j = 1 ; j <= n ; j++)
          {
              if(i==j)dis[i][j]=0;
              else dis[i][j]=INF;
          }
        for (i = 1 ; i <= n ; i++)
            for (j = i+1 ; 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 0 ;
}