2017ccpc网络赛——Friend-Graph

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本文介绍了一个基于拉姆齐定理的算法问题，该问题涉及判断一个由n个成员组成的团队是否为好团队的标准。通过对成员间的朋友关系进行分析，并利用拉姆齐定理简化判断过程，实现了一种高效解决方案。

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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≤3000n≤3000)

Then there are n-1 rows. The iith row should contain n-i numbers, in which number aijaijrepresents 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

Sample Output

Great Team!

题意：一个队里有n个人，告诉你每个人和其他人之前是否是朋友关系，现要你判断这个队是什么队，判断规则：

1.如果有三个或三个以上的人互相不认识就是Bad Team!

2.如果有三个或三个以上的人互相认识就是Bad Team!

3，如果不满足1，2就是Great Team!

思路：

一开始想着先存图然后判断，可是MLE了，仔细一想的确超内存了。。。

后来用vector保存信息，还是MLE（比要求的内存大了几百kb）。。。

再后来就保存每个点的出度入度，暴力判断，结果TLE。。。

比赛最后乱搞一通过了样例就交了，结果WA。。。

最后才知道有一个神奇的定理叫拉姆齐定理：6 个人中至少存在3人相互认识或者相互不认识

这样就可以让大于6（不知道为啥等于6的时候WA，改成大于6就AC）的直接Bad Team，小于等于6的情况暴力就好

orz！

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <queue>
#include <algorithm>
#include <vector>
#include <stack>
#define INF 0x3f3f3f3f
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
int n;
int pl[10][10];
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);
        if(n>6)
        {
            printf("Bad Team!\n");
            continue;
        }
        memset(pl, 0, sizeof(pl));
        for(int i=1; i<n; i++)
        {
            for(int j=i+1; j<=n; j++)
            {
                int x;
                scanf("%d", &x);
                if(x)
                    pl[i][j]=1;
            }
        }
        bool flag=true;
        for(int i=1; i<n; i++)
        {
            for(int j=i+1; j<=n; j++)
            {
                for(int k=j+1; k<=n; k++)
                {
                    if(pl[i][j]&&pl[j][k]&&pl[i][k])
                    {
                        flag=false;
                        break;
                    }
                    if((!pl[i][j]&&!pl[i][k]&&!pl[j][k]))
                    {
                        flag=false;
                        break;
                    }
                }
                if(!flag)
                    break;
            }
            if(!flag)
                break;
        }
        if(flag)
            printf("Great Team!\n");
        else
            printf("Bad Team!\n");
    }
    return 0;
}