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
题解:由于是求异或,我们只需考虑每一位上的 1 的个数即可,比如:对于最高位,如果为偶数,那么小Q只需选一个或者不选即可那么小Q,小T最后该位上的数是相同的,如果为奇数,小Q选一个为必胜,因为小Q就选这一个,剩下都给小T,小T的最终结果必定小于小Q, 因此,我们只需要对每一位上的1的个数加一遍,如果有出现奇数个,则Q必胜,否者平局;
参考代码为:
 
#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
int w[maxn],u[maxn],v[maxn];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0); int T,n;
cin>>T;
while(T--)
{
cin>>n;
bool flag=false;
for(int i=1;i<=n;i++) cin>>w[i];
for(int i=1;i<n;i++) cin>>u[i]>>v[i];
for(int i=0;i<32;i++)
{
int cnt=0;
for(int i=1;i<=n;i++)
{
if(w[i]&1) cnt++;
w[i]>>=1;
}
if(cnt & 1)
{
cout<<"Q"<<endl;
flag=true;
break;
}
}
if(!flag) cout<<"D"<<endl;
} return 0;
}

  

2018HDU多校训练-3-Problem F. Grab The Tree的更多相关文章

  1. 2018 Multi-University Training Contest 3 Problem F. Grab The Tree 【YY+BFS】

    传送门:http://acm.hdu.edu.cn/showproblem.php?pid=6324 Problem F. Grab The Tree Time Limit: 2000/1000 MS ...

  2. Problem F. Grab The Tree HDU - 6324

    题意:给出一棵n个节点的树,每个节点有一个权值,Q和T玩游戏,Q先选一些不相邻的节点,T选剩下的节点,每个人的分数是所选节点的权值的异或和,权值大的胜出,问胜出的是谁. 题解: 话说,这题后面的边跟解 ...

  3. HDU 6324.Problem F. Grab The Tree-博弈(思维) (2018 Multi-University Training Contest 3 1006)

    6324.Problem F. Grab The Tree 题目看着好难,但是题解说的很简单,写出来也很简单.能想出来就是简单的,想不出来就难(讲道理,就算是1+1的题目,看不出来就是难的啊). 和后 ...

  4. 2018HDU多校训练-3-Problem D. Euler Function

    链接:http://acm.hdu.edu.cn/showproblem.php?pid=6322 Problem Description In number theory, Euler's toti ...

  5. 牛客网多校训练第一场 F - Sum of Maximum(容斥原理 + 拉格朗日插值法)

    链接: https://www.nowcoder.com/acm/contest/139/F 题意: 分析: 转载自:http://tokitsukaze.live/2018/07/19/2018ni ...

  6. 2018HDU多校训练-3-Problem M. Walking Plan

    链接:http://acm.hdu.edu.cn/showproblem.php?pid=6331 Walking Plan  Problem Description There are n inte ...

  7. 2018HDU多校训练-3-Problem G. Interstellar Travel

    链接:http://acm.hdu.edu.cn/showproblem.php?pid=6325                                   Interstellar Tra ...

  8. 2018HDU多校训练一 K - Time Zone

    Chiaki often participates in international competitive programming contests. The time zone becomes a ...

  9. 2018HDU多校训练一 D Distinct Values

    hiaki has an array of nn positive integers. You are told some facts about the array: for every two e ...

随机推荐

  1. 重写(OverRide)/重载(Overload)

    方法的重写规则 参数列表必须完全与被重写方法的相同: 返回类型与被重写方法的返回类型可以不相同,但是必须是父类返回值的派生类(java5 及更早版本返回类型要一样,java7 及更高版本可以不同): ...

  2. yum.rpm一点点

    rpm 1.rpm -qi查询包的详细信息 [root@centos7 tmp]# rpm -qi tree Name : tree Version : 1.6.0 Release : 10.el7 ...

  3. Win32窗口消息机制 x Android消息机制 x 异步执行

    如果你开发过Win32窗口程序,那么当你看到android代码到处都有的mHandler.sendEmptyMessage和 private final Handler mHandler = new ...

  4. tcpip协议

    几个概念 1.分层(我们使用四层模型更为贴合我们的实际网络) 分层是为什么,其实和公司中职位是一样的,不同职位的人做不同的事情,然后不同职位的人合起来,一起完成了数据传输的事情. 链路层  在这个层面 ...

  5. JavaScript笔记九

    1.数组方法 reverse() - 可以用来反转一个数组,它会对原数组产生影响 concat() - 可以连接两个或多个数组,它不会影响原数组,而是新数组作为返回值返回 join() - 可以将一个 ...

  6. scrapy结合selenium抓取武汉市环保局空气质量日报

    1.前言 目标网站:武汉市环境保护局(http://hbj.wuhan.gov.cn/viewAirDarlyForestWaterInfo.jspx).scrapy对接selenium模块抓取空气质 ...

  7. 转载-FileZilla Server源码分析(1)

    FileZilla Server源码分析(1) 分类: VC 2012-03-27 17:32 2363人阅读 评论(0) 收藏 举报 serversocketftp服务器usersockets工作 ...

  8. JavaScript的定时器是如何工作的

    理解JavaScript定时器工作原理对于学习JavaScript非常重要.因为JavaScript是单线程运行的,定时器使用场合少,不是很直观.下面通过三个函数来学习JavaScript如何定义,操 ...

  9. 在jsp页面中设置中序号

    首先要使用<c>标签的话需要先引入下面这句话: <%@ taglib prefix="c" uri="http://Java.sun.com/jsp/j ...

  10. 基于JDK1.8的JVM 内存结构【JVM篇三】

    目录 1.内存结构还是运行时数据区? 2.运行时数据区 3.线程共享:Java堆.方法区 4.线程私有:程序计数器.Java 虚拟机栈.本地方法栈 5.JVM 内存结构总结 在我的上一篇文章别翻了,这 ...