NYOJ 545-Metric Matrice【模拟】

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

240 1 2 31 0 1 22 1 0 13 2 1 020 32 0

样例输出

03

来源

第五届河南省程序设计大赛

解题思路：

直接模拟。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int num[30][30];int main(){int t;scanf("%d",&t);while(t--){int i,j;int n;scanf("%d",&n);for(i=0;i<n;i++){for(j=0;j<n;j++){scanf("%d",&num[i][j]);}}int m=1;for(i=0;i<n;i++){if(num[i][i]!=0){printf("1\n");m=0;break;}}if(m==0)continue;for(i=0;i<n;i++){if(m==0)break;for(j=0;j<n;j++){if(i!=j&&num[i][j]<=0){m=0;printf("2\n");break;}}}if(m==0)continue;for(i=0;i<n;i++){if(m==0)break;for(j=0;j<n;j++){if((i!=j)&&(num[i][j]!=num[j][i])) {printf("3\n");m=0;break;}}}if(m==0){continue;}for(i=0;i<n;i++){if(m==0)break;for(j=0;j<n;j++){if(m==0)break;for(int k=0;k<n;k++){if((i!=j)&&(i!=k)&&(j!=k)&&(num[i][j]+num[j][k]<num[i][k])){printf("4\n");m=0;break;}}}}if(m==0)continue;printf("0\n");}return 0;}