F - Grab The Tree HDU - 6324

首先可以发现，若是后手能赢，那么先手取后手的值，一定能赢，所以首先排除后手胜。
对于平局，当且仅当，所有异或值为零，无论怎么取，双方均相等。剩下的便是先手胜利。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#pragma GCC optimize(2)
#define up(i,a,b)  for(int i=a;i<b;i++)
#define dw(i,a,b)  for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
    char ch = getchar(); ll x = 0, f = 1;
    while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
    return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
int n, T;
const int N = 1e5 + 10;
int a[N];
map<int, int>mp;
int main()
{
    T = read();
    while (T--)
    {
        int sum = 0;
        bool flag = 1;
        n = read();
        upd(i, 1, n)a[i] = read(),sum^=a[i];
        int u, v;
        upd(i, 1, n - 1)u = read(), v = read();
        if (sum == 0)printf("D\n");
        else printf("Q\n");
    }
    return 0;
}