HDU6152 Friend-Graph(拉姆齐定理)

Friend-Graph

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1378    Accepted Submission(s): 699

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!

拉姆齐定理：6个人种至少存在3个或3个以上的人互不认识

题意，问给出关系中，是否存在3个或3个以上的人相互认识

所以>=6直接输出bad,<6暴力即可

#include<bits/stdc++.h>#define mem(a,b) memset(a,b,sizeof(a))using namespace std;const int maxn=2e5+10;int e[10][10],n;int main(){    int t,x;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        mem(e,0);        for(int i=1;i<n;i++)        {            for(int j=i+1;j<=n;j++)            {                scanf("%d",&x);                if(x&&n<6)e[j][i]=e[i][j]=1;            }        }        if(n>=6)        {            printf("Bad Team!\n");            continue;        }        bool flag=0;       for(int i=1;i<=n;i++)       {           for(int j=i+1;j<=n;j++)           {               for(int k=j+1;k<=n;k++)               {                   if(e[i][j]&&e[i][k]&&e[j][k])                   {                       flag=true;                       break;                   }               }           }       }        if(!flag)            printf("Great Team!\n");        else            printf("Bad Team!\n");    }}