原题链接
题意：判断图和补图 是否含有三元环

拉姆齐定理

拉姆齐定理：

在>=6个点的完全图中，用红蓝两色染色，一定存在一个红色或者蓝色的三角形。

所有n>=6的话直接输出bad team，否则进行暴力即可。

代码

#include <bits/stdc++.h>
using namespace std;

const int N = 20;
int n, m;
vector<int> g[N];
int vis[N];
int get_sum()
{
    for (int i = 1; i <= n; i++)
    {
        for (int j : g[i])
            if (g[i].size() > g[j].size() || (g[i].size() == g[j].size() && i > j))
                vis[j] = 1;
        for (int j : g[i])
            if (g[i].size() > g[j].size() || (g[i].size() == g[j].size() && i > j))
                for (int k : g[j])
                    if (g[j].size() > g[k].size() || (g[j].size() == g[k].size() && j > k))
                        if (vis[k])
                            return 1;
        for (int j : g[i])
            if (g[i].size() > g[j].size() || (g[i].size() == g[j].size() && i > j))
                vis[j] = 0;
    }
    return 0;
}
void init()
{
    m = 0;
    for (int i = 1; i <= min(10, n); ++i)
        g[i].clear();
}
void solve()
{
    cin >> n;
    init();
    for (int i = 1; i <= n - 1; i++)
    {
        for (int j = i + 1; j <= n; j++)
        {
            int t;
            cin >> t;
            if (t && i < 10 && j < 10)
            {
                m++;
                g[i].push_back(j);
                g[j].push_back(i);
            }
        }
    }
    if (n > 6)
        printf("Bad Team!\n");
    else if (get_sum())
        printf("Bad Team!\n");
    else
        printf("Great Team!\n");
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;

    while (t--)
        solve();
    return 0;
}

暴力题解

注意及时剪枝结束，不然会超时。
代码

#include <bits/stdc++.h>
using namespace std;
// #define int long long
// #define pii pair<int, int>
//***********************必不可少，不然内存超限
#define int short int 
const int N = 3010;
const int mod = 1e9 + 7;
int n, m;
vector<int> g[N];
int vis[N];
int get_sum()
{
    for (int i = 1; i <= n; i++)
    {
        for (int j : g[i])
            if(g[i].size()>g[j].size() || (g[i].size()==g[j].size() && i>j))
                vis[j] = 1;
        for (int j : g[i])
            if(g[i].size()>g[j].size() || (g[i].size()==g[j].size() && i>j))
                for (int k : g[j])
                    if(g[j].size()>g[k].size() || (g[j].size()==g[k].size() && j>k))
                        if (vis[k])
                           return 1;
        for (int j : g[i])
            if(g[i].size()>g[j].size() || (g[i].size()==g[j].size() && i>j))
                vis[j] = 0;
    }
    return 0;
}
void init()
{
    m = 0;
    for (int i = 1; i <= n; ++i)
        g[i].clear();
}
void solve()
{
    cin >> n;
    init();
    for (int i = 1; i <= n - 1; i++){
        for (int j = i + 1; j <= n; j++){
            int t;cin >> t;
            if (t){
                m++;
                g[i].push_back(j);
                g[j].push_back(i);
            }
        }
    }
    int ans1 = 0, ans2 = 0;
    ans1 = get_sum();
    m = n * (n - 1) / 2 - m;
    //反向建边
    for (int i = 1; i <= n; ++i)
    {
        memset(vis, 0, sizeof vis);
        for (int j : g[i])
        {
            vis[j] = 1;
        }
        g[i].clear();
        for (int j = 1; j <= n; j++)
        {
            if (vis[j] == 0)
                g[i].push_back(j);
        }
    }
    ans2 = get_sum();
    if (ans1 == 0 && ans2 == 0)
        cout << "Great Team!" << endl;
    else
        cout << "Bad Team!" << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;

    while (t--)
        solve();
    return 0;
}