NYOJ545-Metric Matrice

Metric Matrice
时间限制：1000 ms  |  内存限制：65535 KB
难度：1
描述

Given as input a square distance matrix, where a[i][j] is the distance between point i and point 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

输入
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).
输出
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

样例输入
2
4
0 1 2 3
1 0 1 2
2 1 0 1
3 2 1 0
2
0 3
2 0
样例输出
0
3
来源
第五届河南省程序设计大赛

 

#include<stdio.h>#include<string.h>int a[50][50];int main(){    int i,j,k,n,m,max=0,flag;    scanf("%d",&n);    while(n--)    {       flag=0;       scanf("%d",&m);       memset(a,0,sizeof(a));       for(i=0;i<m;i++)       for(j=0;j<m;j++)       {          scanf("%d",&a[i][j]);       }       for(i=0;i<m;i++)       for(j=0;j<m;j++)       {          if(a[i][i]!=0)          flag=1;          if(a[i][j]<=0&&i!=j&&flag>2)          flag=2;          if(a[i][j]!=a[j][i]&&flag>3)          flag=3;          for(k=0;k<m;k++)          {             if(i!=j&&j!=k&&i!=k)             {               if(a[i][j]+a[j][k]<a[i][k]&&flag==0)               flag=4;             }          }       }       printf("%d\n",flag);    }    return 0;}