2017CCPC网络选拔赛 1003-Friend-Graph

Friend-Graph

                                                  Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                          Total Submission(s): 6514    Accepted Submission(s): 1610

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

141 1 00 01

Sample Output

Great Team!

用到的 是 六度分离理论：世界上任意六个人中,一定有三个人互相认识,或三个人互相不认识，或者参考百度百科：六度分离理论

注意，这道题目里面， 朋友的朋友不是朋友！！！

代码：

#include <cstdio>#include <cstring>#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;}