博弈题单（一）

CF 1363C Game On Leaves

tag: 博弈论，寄偶性，树上博弈

如果只有一个节点或 \(x\) 就是叶节点，则先手必胜

否则，如果有奇数条边，先手胜。否则后手胜

点击查看代码

#include<bits/stdc++.h>
#define int long long
using namespace std;
using pii=pair<int,int>;
using ll = long long;
using ull = unsigned long long;
const ll inf = 1e18;
const int mod = 998244353;void solve(){int n,x;cin>>n>>x;vector<vector<int>> g(n+1);for(int i=1;i<n;i++){int u,v;cin>>u>>v;g[u].push_back(v);g[v].push_back(u);}if(g[x].size()<=1){cout<<"Ayush\n";return;}if((n-1)&1){cout<<"Ayush\n";}else cout<<"Ashish\n";
}signed main(){ios::sync_with_stdio(0);cin.tie(0);int ct=1;cin>>ct;while(ct--){solve();}
}