题目描述
Ayush and Ashish play a game on an unrooted tree consisting of n nodes numbered 1 to n. Players make the following move in turns:
Select any leaf node in the tree and remove it together with any edge which has this node as one of its endpoints. A leaf node is a node with degree less than or equal to 1.
A tree is a connected undirected graph without cycles.
There is a special node numbered x. The player who removes this node wins the game.
Ayush moves first. Determine the winner of the game if each player plays optimally.
Input
The first line of the input contains a single integer t (1≤t≤10) — the number of testcases. The description of the test cases follows.
The first line of each testcase contains two integers n and x (1≤n≤1000,1≤x≤n) — the number of nodes in the tree and the special node respectively.
Each of the next n−1 lines contain two integers u, v (1≤u,v≤n, u≠v), meaning that there is an edge between nodes u and v in the tree.
Output
For every test case, if Ayush wins the game, print “Ayush”, otherwise print “Ashish” (without quotes).
Examples
input
1
3 1
2 1
3 1
output
Ashish
input
1
3 2
1 2
1 3
output
Ayush
Note
For the 1st test case, Ayush can only remove node 2 or 3, after which node 1 becomes a leaf node and Ashish can remove it in his turn.
For the 2nd test case, Ayush can remove node 2 in the first move itself.
题目大意
Ayush和Ashish玩游戏，他们每次删去一棵无根树的叶节点，最后谁删到x节点，谁就能获胜。（Ayush先手）求最后获胜是谁。
题目分析
首先判断x节点是否为叶结点，如果x就是叶结点那么直接就是Ayush获胜（这是一种特殊情况）。
当n==3时，（2-x-3，x不为叶结点）是保证后手必胜的。
而当n==4时，（保证x不为叶结点）先手取走一个节点。那么此时的状态转变成了：n==3，后手变为先手。那么结果即为先手必胜
依次类推，我们可以发现：
n为偶数时，先手必胜，
n为奇数时，后手必胜。
代码如下
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <map>
#include <unordered_map>
#include <queue>
#include <vector>
#include <set>
#include <algorithm>
#include <iomanip>
#define LL long long
using namespace std;
const int N=1005;
int u[N];
int main()
{
int t;
cin>>t;
while(t--)
{
memset(u,0,sizeof u);
int n,x;
cin>>n>>x;
for(int i=1;i<n;i++) //判断x是否为叶节点
{
int a,b;
cin>>a>>b;
u[a]++,u[b]++;
}
if(u[x]<=1)
{
puts("Ayush");
}
else{ //如果不是，那么n为偶数时，先手必胜。n为奇数时，后手必胜
if(n&1) puts("Ashish");
else puts("Ayush");
}
}
return 0;
}