题目传送：http://acm.hdu.edu.cn/showproblem.php?pid=6152

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：暴搜可以过，因为时间给的多。

然后保存的数组类型不是int是bool，要不然会MLE。

其实也可以发现当点数大于等于6的时候，都是bad team了。

 1 #include <iostream>
 2 #include<bits/stdc++.h>
 3 using namespace std;
 4 
 5 bool a[3002][3002];//评论提醒只需a[7][7]即可
 6 int t,n;
 7 
 8 bool work()
 9 {
10    for(int i=1;i<=n;i++)
11    {
12        for(int j=i+1;j<=n;j++)
13         if(a[i][j]==1)
14         {
15             for(int k=j+1;k<=n;k++)
16                 if (a[j][k]==1 && a[k][i]==1) return 0;
17         } else
18         {
19             for(int k=j+1;k<=n;k++)
20                 if(a[j][k]==0 && a[k][i]==0) return 0;
21         }
22    }
23    return 1;
24 }
25 
26 int main()
27 {
28     scanf("%d",&t);
29     while(t--)
30     {
31         scanf("%d",&n);
32         for(int i=1;i<=n;i++)
33         {
34             for(int j=1;j<=n-i;j++)
35             {
36                 int x;
37                 scanf("%d",&x);
38                 a[i][i+j]=x;
39                 a[i+j][i]=x;
40             }
41         }
42         if (n>6) {printf("Bad Team!\n"); continue;}
43       if(work()) printf("Great Team!\n");
44         else printf("Bad Team!\n");
45     }
46   return 0;
47 }