C. Xor-tree Description 【online|Codeforces #245 (Div. 2)C. Xor-tree（DFS&&贪心】ahub 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?≤?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 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 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 1Sample Output
2
4
7题意
给出一棵树，有n个节点，根为1，每个节点的值不是0就是1，每次可以翻转一个节点，翻转后，当前节点值翻转，其儿子节点不变，孙子追随翻转，依次类推，求最少翻转次数，使得每个节点的值与目标结果相同。
题解：
搜根下面的点 不同就操作改变 (不影响父节点的更新)
AC代码
#include
using namespace std;
#define LL long long
#define CLR(a,b) memset(a,(b),sizeof(a))const int N = 1e5+10;
int s[N], t[N];
int vis[N];
vector v[N];
int n;
int ans;
void dfs(int x,int y,int z)
{
if(vis[x]) return;
vis[x] = 1;
if(s[x]^y != t[x]) {
vis[x] = 2;
ans ++;
y = !y;
}
for(int i = 0;
i >n;
for(int i = 0;
i <= n;
i++) v[i].clear();
int x, y;
for(int i = 1;
i < n;
i++) {
cin>>x>>y;
v[x].push_back(y);
v[y].push_back(x);
}
for(int i = 1;
i <= n;
i++) cin>>s[i];
for(int i = 1;
i <= n;
i++) cin>>t[i];
CLR(vis,0);
ans = 0;
dfs(1,0,0);
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