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 (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1234    Accepted Submission(s): 779

Problem Description

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

 

Source

2018 Multi-University Training Contest 3

题意概括：

给出一棵有 N 个节点的树，给出每个节点的权值。

有Q ，T， D 三个人物。

Q 先在树里面取节点，要求是有边相连的节点只能选其一。

Q选完之后剩下的所有节点都属于 T。

比较 Q 和 T的节点权值异或和，如果 Q 的大则Q赢，反则 T 赢，平局输出 D

解题思路：

根据Q的选取条件，我们知道Q只能在这棵树上隔层取值。

那么我们可以把这棵树的节点分成两堆，第一堆是从第一层开始隔层取，每一层都把该层的节点全部取了。

那么如果这两堆不相等 Q 赢（他先选，选最大的那堆），如果两堆相等 平局。

因为如果出现两堆相等的情况，无论怎么交换节点，两个相同的值异或相同的值结果还是相等。

所以T不可能赢。

BFS跑一遍分成两堆即可。

AC code：

#include <set>
#include <map>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
const int MAXN = 1e5+10;
struct Edge
{
    int v, nxt;
}edge[MAXN<<1];
int head[MAXN], cnt;
int w[MAXN];
int ans_a, ans_b;
void init()
{
    memset(head, -1, sizeof(head));
    ans_a = ans_b = 0;
    cnt = 0;
}

void add(int from, int to)
{
    edge[cnt].v = to;
    edge[cnt].nxt = head[from];
    head[from] = cnt++;
}



void bfs(int x, bool no)
{
    if(no){
        ans_a^=w[x];
        for(int i = head[x]; i != -1; i = edge[i].nxt){
            bfs(edge[i].v, !no);
        }
    }
    else{
        ans_b^=w[x];
        for(int i =head[x]; i != -1; i = edge[i].nxt){
            bfs(edge[i].v, !no);
        }
    }
}

int main()
{
    int N, u, v;
    int T_case;
    scanf("%d", &T_case);
    while(T_case--){
        scanf("%d", &N);
        init();
        for(int i = 1; i <= N; i++){
            scanf("%d", &w[i]);
        }

        for(int i = 1; i < N; i++){
            scanf("%d %d", &u, &v);
            add(u, v);
        }
        ans_a = ans_b = 0;
        bfs(1, true);

        if(ans_a == ans_b) puts("D");
        else puts("Q");

    }
    return 0;
}