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+5int 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;}