nyoj 545 Metric

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

 题目意思就是：判断是否满足所给的4个规则，若满足输出0；反正输出 不满足规则号但须是最小的，比方说 不满足1，3 规则  就要输出较小值1。我的思路：依次判断4个规则。

 
#include<stdio.h>
#include<string.h>
#define max 30+5
int map[max][max];
int main()
{
    int t,n,k,i,j,exist;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        exist=0;
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                scanf("%d",&map[i][j]);
                if(i==j&&map[i][j])
                {
                    exist=1; 
                }
            }
        }
        if(exist)
        {
            printf("%d\n",exist);
            continue;
        }
        for(i=1;i<=n;i++)
        {
            if(exist)//已经违反规则 
            break;
            for(j=1;j<=n;j++)
            {
                if(i!=j&&map[i][j]<=0)
                {
                    exist=2;
                    break;
                }
            }
        }
        if(exist)
        {
            printf("%d\n",exist);
            continue;
        }
        for(i=1;i<=n;i++)
        {
            if(exist)
            break;
            for(j=1;j<=n;j++)
            {
                if(map[i][j]!=map[j][i])
                {
                    exist=3;
                    break;
                }
            }
        } 
        if(exist)
        {
            printf("%d\n",exist);
            continue;
        }
        for(k=1;k<=n;k++)
        {
            if(exist)
            break;
            for(i=1;i<=n;i++)
            {
                if(exist)
                break;
                if(i==k)
                continue;
                for(j=1;j<=n;j++)
                {
                    if(j==k||j==i)
                    continue;
                    if(map[i][k]+map[k][j]<map[i][j])
                    {
                        exist=4;
                        break;
                    }
                }
            }
        }
        printf("%d\n",exist);
    }
    return 0;
}