2017中国大学生程序设计竞赛 - 网络选拔赛 1003 Friend-Graph

本文介绍了一个算法问题，旨在通过分析成员间的友谊关系来判断一个由n名成员组成的团队是否稳定。稳定性判断的标准为：团队中不存在三个及以上成员互相认识或互不认识的情况。文章提供了输入输出示例及AC代码。

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Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3254 Accepted Submission(s): 523

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

i th 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

Sample Output

  
   Great Team!

题意：求是否符合条件，符合认识的队不能>=3,不认识的也不能>=3,符合这两个条件的输出Great Team!,否则输出Band Team!

注意这不是并查集，没有传递性。

ac的代码

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
    int t,n,a[3005],b[3005];
    scanf("%d",&t);
    int i,j,x,k;
    while(t--)
    {
        scanf("%d",&n);
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(i=1; i<=n-1; i++)
        {
            k=1;
            for(j=1; j<=n-i; j++)
            {
                scanf("%d",&x);
                if(x==1)
                {
                    a[i]++;
                    a[i+k]++;
                }
                else
                {
                    b[i]++;
                    b[i+k]++;
                }
                k++;
            }
        }
        bool flag=false;
        for(i=1; i<=n; i++)
        {
            if(a[i]>=3||b[i]>=3)
            {
                flag=true;
                break;
            }
        }
        if(flag)
            printf("Bad Team!\n");
        else
            printf("Great Team!\n");
    }
    return 0;
}