Metric Matrice（nyoj545）

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Metric Matrice

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

简单模拟

被这题给坑了，写下来提醒自己，看清题和注意条件！

代码：

#include<iostream>
#include<cstdio>
#include<string>
#include<string.h>
#include<algorithm>
#include<cmath>
#include<vector>
#include<cstring>
#include<stack>
#include<stdlib.h>
#include<ctype.h>
using namespace std;
#define MAXN 1300
#define inf 100000int graph[35][35],flag[5];
int main()
{int t,n;cin>>t;while(t--){memset(flag,1,sizeof(flag));cin>>n;for(int i=0;i<n;i++)for(int j=0;j<n;j++)cin>>graph[i][j];for(int i=0;i<n;i++){if(graph[i][i]!=0)flag[1]=0;for(int j=0;j<n;j++){if(graph[i][j]<=0&&i!=j)flag[2]=0;if(graph[i][j]!=graph[j][i])flag[3]=0;for(int k=0;k<n;k++){if(graph[i][j]+graph[j][k]<graph[i][k]&&i!=j&&j!=k&&k!=i)//（x）flag[4]=0;}}}bool kl=0;for(int i=1;i<5;i++)if(!flag[i]){cout<<i<<endl;kl=1;break;}if(!kl)cout<<'0'<<endl;}
}

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