Grab The Tree HDU - 6324--博弈

黄油^小饼干

于 2019-05-15 19:29:08 发布

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本文介绍了一款在树形结构上进行的游戏，玩家LittleQ和LittleT轮流选择顶点，目标是在遵循特定规则的情况下获得最大化的异或和得分。文章详细解析了游戏策略，指出最终结果取决于所有顶点价值的异或总和，并提供了实现这一策略的C++代码。

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题目

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

Sample Output

分析

这个题不要被边给迷惑了，跟边没有任何关系。这个题只有两种情况，要么Q赢，要么平局，现在来分析下。根据异或的性质，Q异或T等于全部异或的和，所以Q只要取异或总和最高位为1的那部分，Q的异或和就是最大的。然而如果总的异或和为0，那么Q和T就打成了平局。

代码

#include<iostream>
#include<cstdio>
#include<string.h>
using namespace std;
typedef long long ll;
const int N=1e5+5;
int w[N];
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		int n;
		scanf("%d",&n);
		int ans=0;
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&w[i]);
			ans^=w[i];
		}
		for(int i=1;i<n;i++)
		{
			int a,b;
			scanf("%d%d",&a,&b);
		}
		if(ans)
			printf("Q\n");
		else
			printf("D\n");
	}
	return 0;
}