Friend-Graph

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!

 

 

Source

2017中国大学生程序设计竞赛 - 网络选拔赛

 

题目大意：给定n个人，如果有连续3个人及以上是朋友，或者有3个人及以上不认识，那么Bad Team!，否则Great Team!

#include<iostream>
#include<cstdio>
using namespace std;
int G[3003][3003];
int main()
{
    int T;
	scanf("%d",&T);
	while(T--)
	{	
	    int n;
		scanf("%d",&n);
		for(int i=1;i<=n-1;i++)
			for(int j=i+1;j<=n;j++)			
				scanf("%d",&G[i][j]);
		int flag=0;		
		for(int i=1;i<=n-1;i++)
		{
			int sum1=0,sum2=0;
			for(int j=i+1;j<=n;j++)
			{
				if(G[i][j]==1)
				   sum1++;
				if(G[i][j]==0)
				   sum2++;	
				if(sum1>=3||sum2>=3)
					flag=1;	
			}
	
		}	
		if(flag)
		  printf("Bad Team!\n");		
		else
		  printf("Great Team!\n");		
	}
	
	return 0;
	
}
	

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