下來的第一次相遇是在不翻蓋的同一節(jié)點,遞歸可以是....

A. Xor-tree time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

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 nodex. 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?≤?105). 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?ninteger 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 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 2 4 7

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <vector>using namespace std;int n,init[110000],goal[110000]; vector<int> g[110000],ans;void dfs(int u,int fa,int c1,int c2) {if(c1) init[u]^=1;if(init[u]!=goal[u]){c1^=1; ans.push_back(u);}for(int i=0;i<g[u].size();i++){int v=g[u][i];if(v==fa) continue;dfs(v,u,c2,c1);} }int main() {scanf("%d",&n);for(int i=1;i<n;i++){int a,b;scanf("%d%d",&a,&b);g[a].push_back(b);g[b].push_back(a);}g[0].push_back(1);for(int i=1;i<=n;i++)scanf("%d",init+i);for(int i=1;i<=n;i++)scanf("%d",goal+i);dfs(0,0,0,0);printf("%d\n",(int)ans.size());for(int i=0;i<ans.size();i++)printf("%d\n",ans[i]);return 0; }

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轉載于:https://www.cnblogs.com/mengfanrong/p/4656737.html

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