Graph
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 26 Accepted Submission(s): 14
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.
3
0 1 1
1 0 1
1 1 0
3
0 1 3
4 0 2
7 3 0
3
0 1 4
1 0 2
4 2 0
Case 2: 4
Case 3: impossible
#include<cstdio>
#include<cstring>
int t,tt,n,a[105][105];
int solve()
{
int i,j,k,m=0,f[105][105]={0};
for(k=0;k<n;k++)
for(i=0;i<n;i++)
for(j=0;j<n;j++)
if(i!=j&&i!=k&&k!=j)
{
if(a[i][j]==a[i][k]+a[k][j]&&!f[i][j])
{
f[i][j]=1;
m++;
}
if(a[i][j]>a[i][k]+a[k][j])
return -1;
}
return m;
}
int main()
{
scanf("%d",&t);
while(t--)
{
int i,j,k;
scanf("%d",&n);
for(i=0;i<n;i++)
for(j=0;j<n;j++)
scanf("%d",&a[i][j]);
printf("Case %d: ",++tt);
if((k=solve())==-1)
puts("impossible");
else
printf("%d\n",n*(n-1)-k);
}
}