题目链接:http://nyoj.top/problem/545

  • 内存限制:64MB 时间限制:1000ms

题目描述

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 != j != k

输入描述

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).

输出描述

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

解题思路

判断一下最先违反第几种规则，简单的判断。

#include <bits/stdc++.h>
using namespace std;
int main() {
    int t, n, mp[35][35];
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                scanf("%d", &mp[i][j]);
        int temp = 0;
        for (int i = 0; i < n && !temp; i++) {
            for (int j = 0; j < n && !temp; j++) {
                if (!(i - j) && mp[i][j]) {
                    temp = 1;
                    break;
                }
                else if (i != j && mp[i][j] <= 0) {
                    temp = 2;
                    break;
                }
                else if (mp[i][j] != mp[j][i]) {
                    temp = 3;
                    break;
                }
                else {
                    for (int k = 0; k < n && !temp; k++) {
                        if (i != j && i != k && j != k) {
                            if (mp[i][j] + mp[j][k] < mp[i][k])
                                temp = 4;
                        }
                        else break;    
                    }
                }
            }
        }
        printf("%d\n", temp);
    }
    return 0;
}