Friend-Graph (HDU 6152)2017中国大学生程序设计竞赛

本文介绍了一个算法，用于评估团队是否为好团队。通过检查成员之间的朋友关系，确保没有三人或以上的小团体存在，无论是朋友还是非朋友关系。文章提供了一个C++实现示例，通过遍历成员关系矩阵来判断条件。

Problem Description

It is well known that small groups are not conducive of the development of a team. Therefore, there shouldn’t be any small groups in a good team.
In a team with n members,if there are three or more members are not friends with each other or there are three or more members who are friends with each other. The team meeting the above conditions can be called a bad team.Otherwise,the team is a good team.
A company is going to make an assessment of each team in this company. We have known the team with n members and all the friend relationship among these n individuals. Please judge whether it is a good team.

Input

The first line of the input gives the number of test cases T; T test cases follow.（T<=15）
The first line od each case should contain one integers n, representing the number of people of the team.(n≤3000)

Then there are n-1 rows. The ith row should contain n-i numbers, in which number aij represents the relationship between member i and member j+i. 0 means these two individuals are not friends. 1 means these two individuals are friends.

Output

Please output ”Great Team!” if this team is a good team, otherwise please output “Bad Team!”.

Sample Input

1 4 1 1 0 0 0 1

Sample Output

Great Team!

题解：Team中有3个或者3个以上不是朋友或者是朋友，都输出Bad，其他输出Great。

/** By Mercury_Lc */
#include <bits/stdc++.h>
using namespace std;
int a[3005][3005];
int main()
{
    int t,i,j,k,n,f;
    scanf("%d",&t);
    while(t--)
    {
        f = 0;
        scanf("%d",&n);
        for(i = 1; i <= n; i ++)
        {
            for(j = i + 1; j <= n; j ++)
            {
                scanf("%d",&a[i][j]);
            }
        }
        for(i = 1; i <= n; i ++)     //暴力搜索
        {
            for(j = i + 1; j <= n; j ++)
            {
                if(a[i][j])          // 是朋友关系，继续判断这两个人有没有共同朋友
                {
                    for(k = j + 1; k <= n; k ++)
                    {
                        if(a[i][k] && a[j][k])
                        {
                            f = 1;
                            break;
                        }
                    }
                }
                else           // 不是朋友关系，继续判断这两个人有没有共同朋友
                {
                    for(k =  j + 1; k <= n; k ++)   
                    {
                        if(a[i][k] == 0 && a[j][k] == 0)
                        {
                            f = 1;
                            break;
                        }
                    }
                }
                if(f)break;
            }
            if(f)break;
        }
        if(f)printf("Bad Team!\n");
        else printf("Great Team!\n");
    }
    return 0;
}

转载于:https://www.cnblogs.com/lcchy/p/10139587.html