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总共用2^n中可能，但是我tm的，就想着贪心了，woc，真的啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊。
Problem - D - Codeforces
题解地址
思路，对于n个牌子，有相同的牌子必然是不能再一起的。这里想到二分图。用染色法判断二分图。
E. Split Into Two Sets（ranting 1600）超详细题解-CSDN博客
#include <bits/stdc++.h>using namespace std;
typedef long long ll;
typedef double db;
typedef long double ldb;
typedef pair<int, int> pii;
typedef pair<ll, ll> PII;
#define pb emplace_back
//#define int ll
#define all(a) a.begin(),a.end()
#define x first
#define y second
#define ps push_back
#define endl '\n'
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define lc u << 1
#define rc u << 1 | 1void solve();const int N = 1e6 + 10;signed main() {IOS;ll t = 1;cin >> t;while (t--)solve();return 0;
}struct eage{ll v,ne;}e[N];
ll h[N],idx;
ll color[N];//2种颜色，1和2ll n;
void add(ll u,ll v)
{e[++idx] = {v, h[u]};h[u] = idx;
}bool dfs(ll u,ll c)//节点，颜色
{color[u] = c;//染色for(int i = h[u]; i; i = e[i].ne){ll v = e[i].v;if(!color[v]){if(dfs(v,3-c)) return 1;//有奇环}else if(color[v] == c) return 1;//颜色相同，有奇环}return 0;
}void init()
{for(int i = 0; i <= idx; ++ i){e[i] = {0,0};}idx = 0;for(int i = 1; i <= n; ++ i){h[i] = color[i] = 0;}
}void solve() {init();//多测不初始化，亲人2行泪。wa1，wa2，wa3，，，cin >> n;vector<ll> num(n+1,0);bool ok = 1;vector<vector<ll>> ee(n+1);for(int i=1,u,v; i <= n; ++ i){cin >> u >> v;++ num[u],++ num[v];ee[u].ps(i),ee[v].ps(i);if(u == v) ok = 0;}for(int i = 1; i <= n; ++ i)if(num[i]>2 || ok == 0){cout << "NO" << endl;return;}for(int i = 1; i <= n; ++ i){ll u,v;u = ee[i][0],v = ee[i][1];add(u,v);add(v,u);}for(int i = 1; i <= n; ++ i){if(!color[i]){if(dfs(i,1)){ok = 0;//有奇环break;}}}for(int i = 1; i <= n; ++ i){if(!color[i]){ok = 0; break;}}if(ok) cout << "YES" << endl;else cout << "NO" << endl;
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