HDUoj 6152 Friend-Graph ( 数学/暴力_it is well known that small groups are not conduci-优快云博客

本文链接：https://blog.youkuaiyun.com/wang2332/article/details/77418403

此算法通过检查团队成员间的友谊关系来判断一个由 n 个人组成的团队是否良好。良好的团队不能存在三人以上的小团体或者三人以上互不相识的情况。代码采用暴力搜索法检查所有可能的三人组合。

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Friend-Graph

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

Sample Output

Great Team!

题意

找一个完全图里面的三元环直接暴力就好（卡内存要用bool数组
不过网赛打完dalao们发了一个定理拉姆齐定理（还好出题人没有丧心病狂卡这个定理不然不知道的肯定过不了了

https://baike.baidu.com/item/%E6%8B%89%E5%A7%86%E9%BD%90%E5%AE%9A%E7%90%86/7478185?fr=aladdin

AC代码

#include <bits/stdc++.h>
using namespace std;
int arr[10][10];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--) {
        int n;
        scanf("%d",&n);
        int x;
        if(n >= 6) {
            for(int i = 1; i <= n; i++) {
                for(int j = i+1; j <= n; j++) scanf("%d",&x);
            }
            puts("Bad Team!"); continue;
        }
        for(int i = 1; i <= n; i++) {
            for(int j = i+1; j <= n; j++) scanf("%d",&arr[i][j]);
        }
        bool flag = false;
        for(int i = 1; i <= n; i++) {
            for(int j = i+1; j <= n; j++) {
                for(int k = j+1; k <= n; k++) {
                    if(arr[i][j] && arr[j][k] && arr[i][k]) flag = true;
                    if(!arr[i][j] && !arr[j][k] && !arr[i][k]) flag = true;
                }
            }
        }
        if(flag) puts("Bad Team!");
        else puts("Great Team!");
    }
return 0;
}