nyoj 545 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

#include<stdio.h>
#include<string.h>
int x[35][35];
void slove(int n)
{
	for(int i=0;i<n;++i) 
	{
		if(x[i][i]!=0)
		{
			printf("1\n");
			return;
		}
	}
	for(int i=0;i<n;++i)
	{
		for(int j=0;j<n;++j)
		{
			if(i!=j&&x[i][j]<=0)
			{
				printf("2\n");
				return;
			}
		}
	}
	for(int i=0;i<n;++i)
	{
		for(int j=0;j<n;++j)
		{
			if(i!=j&&x[i][j]!=x[j][i])
			{
				printf("3\n");
				return;
			}
		}
	}
	for(int i=0;i<n;++i)
	{
		for(int j=0;j<n;++j)
		{
			for(int k=0;k<n;++k)
			{
				if(i!=j&&j!=k&&i!=k&&x[i][j]+x[j][k]<x[i][k])
				{
					printf("4\n");
					return ; 
				}
			}
		}
	}
	printf("0\n");
}
int main()
{
	int t,n;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(int i=0;i<n;++i)
		{
			for(int j=0;j<n;++j)
			{
				scanf("%d",&x[i][j]);
			}
		}
		slove(n);
	}
	return 0;
}