The shortest path between any two points (Floyed)

F、Moving On

Firdaws and Fatinah are living in a country with

Firdaws's home locates in the city

Input

The input contains several test cases, and the first line is a positive integer

For each test case, the first line contains two integers $\le n \le 200)n(1≤n≤200) which is the number of cities, and q (1 \le q \le 2 \times 10^4)q(1≤q≤2×104) which is the number of queries that will be given. The second line contains nn integers r_1, r_2, \cdots , r_nr1,r2,⋯,rn indicating the risk of kidnapping or robbery in the city 11 to nn respectively. Each of the following nn lines contains nn integers, the jj-th one in the ii-th line of which, denoted by d_{i,j}di,j, is the distance from the city iito the city jj.$

Each of the following

We guarantee that $\le r_i \le 10^5, 1 \le d_{i,j} \le 10^5 (i \neq j), d_{i,i} = 01≤ri≤105,1≤di,j≤105(i=j),di,i=0 and d_{i,j} = d_{j,i}di,j=dj,i. Besides, each query satisfies 1 \ and u, v \ n 1 ≤ you , v ≤ n and 1 \ and w \ 10 ^ 5 1 ≤ w ≤ 1 0 5.$

Output

For each test case, output a line containing Case #x: at first, where

Extra spaces at the end of each row, the answer does not affect the validity of the output

Sample input

Sample Output

#. 1 Case: 
0 
. 1 
. 3 
0 
. 1 
2 

that Italy: an input t, t represents a test sample 
input n, q, n represents points, q interrogation times 
the number of the next row and n represents a dangerous [1, n] point value 
next n lines, each line number 3 is described an edge, from the point x to y to z; 
last q rows of three numbers, starting from the seek to the end U V, does not exceed the value at each point dangerous the shortest distance under the premise of W

#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
#include<math.h>
#include<string>
#include<string.h>
#include<vector>
#include<utility>
#include<map>
#include<queue>
#include<set>
#define mx 0x3f3f3f3f
#define ll long long
#define MAXN 100
using namespace std;
intDP [ 250 ] [ 250 ] [ 250 ]; // DP [K] [I] [J] represents added (after 1 ~ k Urban small risk value) is i-> j shortest 
int VIS [ 250 ] , RK [ 250 ];
 BOOL CMP ( int I, int J) 
{ 
    return RK [I] < RK [J]; 
} 
int main () 
{ 
    int T; 
    Scanf ( " % D " , & T);
     for ( int = TT . 1 ; TT <= T; TT ++ ) 
    { 
        Memset (DP, MX, the sizeof(DP));
         int n-, Q; 
        Scanf ( " % D% D " , & n-, & Q);
         for ( int I = . 1 ; I <= n-; I ++ ) 
        { 
            VIS [I] = I; // initialize rank 
            Scanf ( " % D " , & RK [I]); 
        } 
        Sort (VIS + . 1 , VIS + n-+ . 1 , CMP); // the city numbered level of risk from small to large, vis [ranking] = number
         // for (int I =. 1; I <= n-; I ++)
         //      COUT << VIS [I] << '';
         //<< COUT "<-----" << endl; 
        for ( int I = . 1 ; I <= n-; I ++ )
             for ( int J = . 1 ; J <= n-; J ++ ) 
                Scanf ( " % D " , & DP [ 0 ] [I] [J]); // hazard level initialize each edge is 0 
        for ( int K = . 1 ; K <= n-; K ++ ) 
        { 
            int now = VIS [K];
             for ( int I = . 1 ; I <= n-; I ++ ) 
            { 
                for ( int J =1;j<=n;j++)
                    dp[k][i][j]=min(dp[k-1][i][j],dp[k-1][i][now]+dp[k-1][now][j]); 
            }

        }
        printf("Case #%d:\n",tt);
        for(int i=1;i<=q;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            int k=0;
            for(int j=1; J <= n-; J ++ )
                 IF (RK [VIS [J]] <= w) // shortest side after the risk value closest to w are added to the list 
                    K = J; 
            the printf ( " % D \ n- " , DP [K] [U] [V]); 
        } 
    } 
    return  0 ; 
}