hdu6152 Friend-Graph （2017 ccpc 网络赛 （暴力 + 一个神奇的定理）

Friend-Graph

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 995    Accepted Submission(s): 512

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 

 

Recommend

liuyiding

 

  思路：Ramsey theorem（ 拉姆齐定理 ），  n 个人中必定有 k 个人相识或 k 个人互不相识 n = 6，k = 3；

#include 
#include 
#define maxn 6  //只考虑 n= 3,4,5 的情况即可，暴力
using namespace std;

int a[maxn][maxn];  //邻接矩阵 存储无向图
int main()
{
    int T,n;
    scanf("%d",&T);
    while(T--)
    {
        int flag=0;
        scanf("%d",&n);
        if(n>=6)
            flag=1;
        if(n==1 || n==2)
            flag=2;

        memset(a,0,sizeof a);
        int value;
        for(int i=1;i<=n-1;i++) {
            for(int j=1;j<=n-i;j++)
            {
                scanf("%d",&value);
                a[i][i+j]=1;
                a[j+i][i]=1;
            }
        }
        if(flag == 2)
            printf("Great Team!\n");
        else if(flag == 1)
            printf("Bad Team!\n");
        else if(flag == 0){
            for(int i=1;i<=n;i++) {
               for(int j=1;j<=n;j++) {
                  for(int k=1;k<=n;k++)
                  {
                      if(i==k || i==j || j==k)
                        continue;
                      //下面是 符合题目中说的， Bad 的两种情况
                      if(a[i][j]==1&&a[i][k]==1&&a[j][k]==1)  {  flag=1; break;  }
                      if(a[i][j]==0&&a[i][k]==0&&a[j][k]==0)  {  flag=1; break;  }
                  }
                  if(flag) break;
                }
                if(flag) break;
            }
            if(flag == 1)  printf("Great Team!\n");
            else    printf("Bad Team!\n");
        }
    }
    return 0;
}