1478: Metric Matrice
Time Limit: 1 Sec
Memory Limit: 128 MB
Submit: 70
Solved: 35
Submit
Status
Web Board
Description
nt j, determine if the distance matrix is "a metric" or not.
A distance matrix a[i][j] is a metric if and only if
1. a[i][i] = 0
2. a[i][j]> 0 if i != j
3. a[i][j] = a[j][i]
4. a[i][j] + a[j][k] >= a[i][k] i ¹ j ¹ k
Input
The first line of input gives a single integer, 1 ≤ N ≤ 5, the number of test cases. Then follow, for each test case,
* Line 1: One integer, N, the rows and number of columns, 2 <= N <= 30
* Line 2..N+1: N lines, each with N space-separated integers
(-32000 <=each integer <= 32000).
Output
Output for each test case , a single line with a single digit, which is the lowest digit of the possible facts on this list:
* 0: The matrix is a metric
* 1: The matrix is not a metric, it violates rule 1 above
* 2: The matrix is not a metric, it violates rule 2 above
* 3: The matrix is not a metric, it violates rule 3 above
* 4: The matrix is not a metric, it violates rule 4 above
Sample Input
2
4
0 1 2 3
1 0 1 2
2 1 0 1
3 2 1 0
2
0 3
2 0
Sample Output
0
3
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[31][31];
int main()
{
int t,n,i,j,k;
scanf("%d",&t);
while(t--)
{
memset(a,0,sizeof(a));
scanf("%d",&n);
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
scanf("%d",&a[i][j]);
}
}
int flag=0;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
if(i==j&&a[i][j]!=0)
{
flag=1;
break;
}
else if(i!=j&&a[i][j]<=0)
{
flag=2;
break;
}
else if(a[i][j]!=a[j][i])
{
flag=3;
break;
}
else
{
for(k=0;k<n&&k!=i&&k!=j&&i!=j;k++)
{
if(a[i][j]+a[j][k]<a[i][k])
{
flag=4;
break;
}
}
if(flag)
break;
}
}
if(flag)
break;
}
printf("%d\n",flag);
}
return 0;
}
Source
第五届河南省大学生程序设计竞赛
