Codeforces Round #245 (Div. 2) C. Xor-tree DFS

C. Xor-tree

Time Limit: 1 Sec Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/430/problem/C

Description

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 init_i, 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 goal_i, 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 ≤ 10⁵). Each of the next n - 1 lines contains two integers u_i and v_i (1 ≤ u_i, v_i ≤ n; u_i ≠ v_i) meaning there is an edge between nodes u_i and v_i.

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

Output

Sample Input

10
1
1
2
1
2
5
6
8
5
0 1 1 0 1 0 1 0 1
0 1 0 0 1 1 1 0 1

Sample Output

2
4
7

HINT

题意

给你一棵以1为根节点的树，然后每个点都是1或者0,

你可以使得一个点和他的儿子们全部都翻转

然后问你最少翻转多少次，且翻转的是哪些点

题解：

很明显的一个dfs

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
//const int inf=0x7fffffff;   //无限大
const int inf=0x3f3f3f3f;
/*

int buf[10];
inline void write(int i) {
  int p = 0;if(i == 0) p++;
  else while(i) {buf[p++] = i % 10;i /= 10;}
  for(int j = p-1; j >=0; j--) putchar('0' + buf[j]);
  printf("
");
}
*/
//**************************************************************************************
inline ll read()
{
    int x=0,f=1;char ch=getchar();
    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;
}
struct edge
{
    int x,y,z;
};
vector<int> e[maxn];
void add_edge(int a,int b)
{
    e[a].push_back(b);
}
int a[maxn];
int dp[maxn];
vector<int> ans;
int flag[maxn];
void dfs(int x,int c,int d)
{
    if(flag[x])
        return;
    flag[x]=1;
    if(a[x]^c==dp[x])
    {
        for(int i=0;i<e[x].size();i++)
        {
            int v=e[x][i];
            dfs(v,d,c);
        }
    }
    else
    {
        ans.push_back(x);
        for(int i=0;i<e[x].size();i++)
        {
            int v=e[x][i];
            dfs(v,d,!c);
        }
    }
}
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n-1;i++)
    {
        int u=read(),v=read();
        add_edge(v,u);
        add_edge(u,v);
    }
    for(int i=1;i<=n;i++)
        cin>>a[i];
    for(int i=1;i<=n;i++)
        cin>>dp[i];
    dfs(1,0,0);
    cout<<ans.size()<<endl;
    for(int i=0;i<ans.size();i++)
        cout<<ans[i]<<endl;
}