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();
    }
}