Codeforces Round #245 (Div. 1) A /(Div.2 C) Xor-tree

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原题链接： http://codeforces.com/contest/429/problem/A

Xor-tree

Iahub is very proud of his recent discovery, propagating trees. Right now, he invented a new tree, called xor-tree. After this new revolutionary discovery, he invented a game for kids which uses xor-trees.

The game is played on a tree having n nodes, numbered from 1 to n. Each node i has an initial value initi, which is either 0 or 1. The root of the tree is node 1.

One can perform several (possibly, zero) operations on the tree during the game. The only available type of operation is to pick a node x. Right after someone has picked node x, the value of node x flips, the values of sons of x remain the same, the values of sons of sons of x flips, the values of sons of sons of sons of x remain the same and so on.

The goal of the game is to get each node i to have value goali, which can also be only 0 or 1. You need to reach the goal of the game by using minimum number of operations.

Input

The first line contains an integer n (1 ≤ n ≤ 10^5^). Each of the next n - 1 lines contains two integers ui and vi (1 ≤ ui, vi ≤ n; ui ≠ vi) meaning there is an edge between nodes ui and vi.

The next line contains n integer numbers, the i-th of them corresponds to initi (initi is either 0 or 1). The following line also contains n integer numbers, the i-th number corresponds to goali (goali is either 0 or 1).

Output

In the first line output an integer number cnt, representing the minimal number of operations you perform. Each of the next cnt lines should contain an integer xi, representing that you pick a node xi.

Sample test(s)

Input

1
2
3
4
5
6
7
8
9
10
11
12
10
2 1
3 1
4 2
5 1
6 2
7 5
8 6
9 8
10 5
1 0 1 1 0 1 0 1 0 1
1 0 1 0 0 1 1 1 0 1

Output

1
2
3
2
4
7

这道题我自己做是没有任何思路的，无奈之下研究了大神的代码，解法实在是太漂亮了，我这种菜鸟想不到是很正常的！

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#include <cstdio>
#include <vector>
#define maxn 100010

std::vector<int> Edge[maxn];
int n, a[maxn], b[maxn];
int cnt = 0, ans[maxn];

void dfs(int rt, int pre, int p1, int p2){
    if( a[rt]^ p1 != b[rt] ){      //(1/0)与0异或不翻转；与1异或 翻转
        ans[cnt++] = rt;    //a、b不同需要 flip
        p1 = 1- p1;    //翻转 flip
    }

    for(int i = 0; i < Edge[rt].size(); i++){
        int &e = Edge[rt][i];     //注意此处的引用
        if(e == pre) continue;    //父节点
        dfs(e, rt, p2, p1);     //p1 、p2参数位置的交换实现flip的“隔代遗传”
    }
}

int main(){
    int i, x, y;
    scanf("%d", &n);
    for(i = 1; i < n; i++){
       scanf("%d %d",&x, &y);
       Edge[x].push_back(y);
       Edge[y].push_back(x);
    }
    for(i = 1; i <= n; i++)  scanf("%d", a + i);
    for(i = 1; i <= n; i++)  scanf("%d", b + i);
    dfs(1, -1, 0, 0);
    printf("%d\n", cnt);
    for(i = 0; i < cnt; i++)
        printf("%d\n", ans[i]);
    return 0;
}