Problem Description

Everyone knows how to calculate the shortest path in a directed graph. In fact, the opposite problem is also easy. Given the length of shortest path between each pair of vertexes, can you find the original graph?

 

Input

The first line is the test case number T (T ≤ 100).
First line of each case is an integer N (1 ≤ N ≤ 100), the number of vertexes.
Following N lines each contains N integers. All these integers are less than 1000000.
The jth integer of ith line is the shortest path from vertex i to j.
The ith element of ith line is always 0. Other elements are all positive.

 

Output

For each case, you should output “Case k: ” first, where k indicates the case number and counts from one. Then one integer, the minimum possible edge number in original graph. Output “impossible” if such graph doesn't exist.

 

Sample Input

 

Sample Output

 

 #include<cstdio>

 #include<iostream>

 #include<cstring>

 using namespace std;

 #define N 103

 #define INF 0x1fffffff

 int mp[N][N];

 int dist[N][N];

 int main()

 {

     int i , j , k ;

     int n;

     int t ;

     cin>>t;

     int c = ;

     while(t--)

     {

         c++;

         scanf("%d",&n);

         for( i =  ; i < n ;i++)

         {

             for(  j =  ; j < n ;j++)

             {

                 scanf("%d",&mp[i][j]);

                 dist[i][j] = mp[i][j];

             }

         }

         bool flag = true;

         for(k =  ;flag && k < n ; k++)

         {

             for(i =  ;flag && i < n ; i++)

             {

                 for( j =  ; flag&& j < n ;j++)

                 {

                     if(dist[i][j]>dist[i][k]+dist[k][j])

                         flag = false;

                 }

             }

         }

         int cnt = ;

         if(flag)

         {

             for( k =  ; k < n ;k++)

             {

                 for(i =  ; i < n ;i++)

                 {

                     for(j =  ;j < n ;j++)

                     {

                         if(i==j||j==k||k==i) continue;

                         if(dist[i][j]==dist[i][k]+dist[k][j])

                             {

                                 dist[i][j] = INF;

                                 //printf("%d %d %d\n", k ,i , j);

                                 cnt++;

                             }

                     }

                 }

             }

         }

         if(flag) printf("Case %d: %d\n",c,n*(n-)-cnt);

         else printf("Case %d: impossible\n",c);

     }

     return  ;

 }