郑轻 1478 Metric Matrice （第五届河南省赛）

1478: Metric Matrice

Time Limit: 1 Sec Memory Limit: 128 MB

Submit: 72 Solved: 37

SubmitStatusWeb Board

Description

nt 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

Input

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).

Output

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

Sample Input

2

4

0 1 2 3

1 0 1 2

2 1 0 1

3 2 1 0

2

0 3

2 0

Sample Output

0

3

HINT

Source

第五届河南省大学生程序设计竞赛

题解：这道题是说：给出一个矩阵，看是否是符合规则的。如果均满足下列四种规则，输出0.

如果不满足1输出1

如果不满足2输出2

如果不满足3输出3

如果不满足4输出4

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>

using namespace std;
const int maxx=1005;
int a[105][105];
int n;
int x1,x2,x3,x4;
///是否符合第一种
inline int f1()
{
     x1=1;
    for(int i=0;i<n;i++)
    {
        if(a[i][i]!=0)
            {
                x1=0;
               return x1;
            }

    }
    return x1;
}
///是否符合第二种
inline int f2()
{
    x2=1;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        {
            if(i!=j)
            {
              if(a[i][j]<=0)
                {
                    x2=0;
                  return x2;
                }
            }
        }
    }
    return x2;
}
///是否符合第三种
inline int f3()
{
    x3=1;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        {
            if(a[i][j]!=a[j][i])

                {
                    x3=0;
                    return x3;
                }

        }
    }
    return x3;
}
///是否符合第四种
inline int f4()
{
    x4=1;
   for(int i=0;i<n;i++)
   {
       for(int j=0;j<n;j++)
       {
           for(int k=0;k<n;k++)
           {
               if(i!=j&&i!=k&&j!=k)
               if(a[i][j]+a[j][k]<a[i][k])
               {
                   x4=0;
                   return x4;
               }
           }
       }
   }
   return x4;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                scanf("%d",&a[i][j]);
            }
        }
        int a1=f1();
        int a2=f2();
        int a3=f3();
        int a4=f4();
        if(a1==1&&a2==1&&a3==1&&a4==1)
            printf("0\n");
            else
                if(a1==0)
                printf("1\n");
        else
            if(a2==0)
            printf("2\n");
        else
            if(a3==0)
            printf("3\n");
        else
            if(a4==0)
            printf("4\n");

    }

    return 0;
}