Codeforces Round #588 (Div. 2) C. Anadi and Domino

https://codeforces.com/contest/1230
直接搜 主要是题解的方法太秀了 居然还能想到 找直接冲突点 然后减掉 找最大值

下面是直接搜索的 枚举每个点可以是啥 6*6的乱搞

#include <bits/stdc++.h>
using namespace std;
const int maxn = 50;

int n, m, ans;
int x[maxn], y[maxn];
int val[maxn];
bool vis[maxn][maxn];

void dfs(int pos) {
    if(pos > n) {
        int tmp = 0;
        memset(vis, 0, sizeof vis);
        for(int i = 1; i <= m; i++) {
            int a = val[x[i]], b = val[y[i]];
            if(!vis[a][b]) {
                tmp ++;
                vis[a][b] = vis[b][a] = 1;
            }
        }
        ans = max(tmp, ans);
        return ;
    }
    for(int i = 1; i <= 6; i++) {
        val[pos] = i;
        dfs(pos + 1);
    }
}

int main() {
    cin >> n >> m;
    // 题解也是nb啊 n <= 6的时候就是m个 6个完全够放不同的 最多15边
    for(int i = 1, a, b; i <= m ;i ++)
        cin >> x[i] >> y[i];
    dfs(1);
    cout << ans <<endl;
	return 0;
}

#include <bits/stdc++.h>
using namespace std;
//#define int long long
const int maxn = 2e5 + 5;
const long long INF = 0x3f3f3f3f3f3f3f3f;
typedef long long ll;
typedef pair<int, int> P;
int n, m, k, cas;
int mp[15][15];
 
signed main() {
	cin >> n >> m;
	for(int i = 1, a, b; i <= m; i ++) {
		cin >> a >> b;
		mp[a][b] = mp[b][a] = 1;
	}
	if(n <= 6) cout << m << endl;
	else {
		int ans = 1000;
		for(int i = 1; i <= n; i ++) {
			for(int j = i + 1; j <= n;  j++) {
				int tmp = 0;
				for(int k = 1; k <= n; k ++) {
					if(mp[i][k] && mp[j][k]) tmp ++;
				}
				ans = min(tmp, ans);
			}
		}
		cout << m - ans << endl;
	}
	return 0;
}