Problem F. Grab The Tree博弈

Little Q and Little T are playing a game on a tree. There are n vertices on the tree, labeled by 1,2,…,n, connected by n−1 bidirectional edges. The i-th vertex has the value of wi.
In this game, Little Q needs to grab some vertices on the tree. He can select any number of vertices to grab, but he is not allowed to grab both vertices that are adjacent on the tree. That is, if there is an edge between x and y, he can’t grab both x and y. After Q’s move, Little T will grab all of the rest vertices. So when the game finishes, every vertex will be occupied by either Q or T.
The final score of each player is the bitwise XOR sum of his choosen vertices’ value. The one who has the higher score will win the game. It is also possible for the game to end in a draw. Assume they all will play optimally, please write a program to predict the result.
Input
The first line of the input contains an integer T(1≤T≤20), denoting the number of test cases.
In each test case, there is one integer n(1≤n≤100000) in the first line, denoting the number of vertices.
In the next line, there are n integers w1,w2,…,wn(1≤wi≤109), denoting the value of each vertex.
For the next n−1 lines, each line contains two integers u and v, denoting a bidirectional edge between vertex u and v.
Output
For each test case, print a single line containing a word, denoting the result. If Q wins, please print Q. If T wins, please print T. And if the game ends in a draw, please print D.
Sample Input
1
3
2 2 2
1 2
1 3
Sample Output
Q

大体意思就是说Q和T两个人玩一个游戏，有n个点每个点有一个权值，Q选一部分不相邻的点，T拿走剩下全部点，两个人选取的点的位数的异或和大的胜利。看似给的数据很多，实际上边是完全用不到的。输入时对所有权值求异或和，如果异或和为0说明所有值相等，两个人最后一定平局，输出D即可，在不为零的情况下，选取异或和最高位为1的那一位，不难的出一定有奇数个这一位为1的权值，只要Q先拿走一个这样的点，那么剩下的在T拿走后异或值一定为0，肯定是Q胜利，所以得到很简洁的代码。顺便心疼T，怎么也赢不了。

#include<bits/stdc++.h>
using namespace std;
int main()
{int T;scanf("%d",&T);while(T--){int a,t;scanf("%d",&a);int ans=0;for(int i=0;i<a;i++){scanf("%d",&t);ans^=t;}for(int i=;i<a-1;i++)scanf("%d%d",&t,&t);if(ans==0) printf("D\n");else printf("Q\n");}return 0;
}

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