Hdu 6152 Friend-Graph（Ramsey定理）

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
这道题在比赛的时候wa了一次，当时对边界情况判断少了n的情况，然后就是卡内存，直接用bool，比赛的时候因为是签到题但时间给了10秒，就是一个暴力题，然后评测机就爆炸了….先给出自己暴力的代码：

#include<iostream>#include <cstring>using namespace std;using LL=int64_t;const int INF=0x3f3f3f3f;bool ans[3005][3005];int main(){    int T;    cin>>T;    while(T--) {        memset(ans,0,sizeof(ans));        int n;        cin>>n;        for(int i=1;i<n;i++) {            for(int j=i+1;j<=n;j++) {                cin>>ans[i][j];                ans[j][i]=ans[i][j];            }        }        int 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(ans[i][j]==ans[i][k]&&ans[i][j]==ans[j][k]) {                        flag=1;                        break;                    }                }                if(flag==1) break;            }            if(flag==1) break;        }        if(flag==0) cout<<"Great Team!"<<endl;        else cout<<"Bad Team!"<<endl;    }    return 0;}

但这道题不仅是用暴力的代码，还可以用Ramsey直接秒掉，这个公式的内容就是：任意六个人中要么至少三个人认识，要么至少三个不认识
看到这个公式的我第一眼就觉得这是我这辈子最后一次见到这个公式了…如果n>=6就直接退出，否则暴力解