第五届省赛题 Metric Matrice

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

题意：
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

#include<stdio.h>#include<iostream>#include<string.h>#include<algorithm>using namespace std;int map[35][35];int n;int f1(){    int a1=1;    for(int i=1;i<=n;i++)    {        if(map[i][i]!=0)        {            a1=0;            return a1;        }    }    return a1;}int f2(){    int a2=1;    for(int i=1;i<=n;i++)    {        for(int j=1;j<=n;j++)        {            if(i!=j)                if(map[i][j]<=0)                {                    a2=0;                    return a2;                }        }    }    return a2;}int f3(){    int a3=1;    for(int i=1;i<=n;i++)        for(int j=1;j<=n;j++)        {            if(map[i][j]!=map[j][i])            {                a3=0;                return a3;            }        }        return a3;}int f4(){    int a4=1;    for(int i=1;i<=n;i++)        for(int j=1;j<=n;j++)            for(int k=1;k<=n;k++)            {                if(i!=j&&i!=k&&j!=k)                {                    if(map[i][j]+map[j][k]<map[i][k])                    {                        a4=0;                        return a4;                    }                }            }        return a4;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        int i,j;        scanf("%d",&n);        for(i=1;i<=n;i++)            for(j=1;j<=n;j++)            {                scanf("%d",&map[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;}