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

原题链接

C. 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.
【Codeforces Round #245 (Div. 2)-C. Xor-tree】 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 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.
Examples 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

节点初始标志记在d1[]中，节点目标标志记在d2[]中，k1为1表示奇数节点标志改变，k2为1表示偶数节点标志改变.h表示节点深度
从根节点开始深搜，遇到节点j先通过k1, k2改变d1[j]的值，改变后若d1[j] != d2[j]则d1[j] ^= 1, 同时若该节点深度h为奇数，则该节点孩子中奇数深度的节点都要改变，所以k1 ^= 1,同理若h为偶数，则该节点孩子中偶数深度的节点都要改变，所以k2 ^= 1;

#include #include #include #include #include #include #include #define maxn 100005 #define MOD 1000000007 #define INF 1e9 using namespace std; typedef long long ll; vector v[maxn], ans; int d1[maxn], d2[maxn]; void dfs(int j, int f, int k1, int k2, int h){ if(k1 && (h&1)){ d1[j] ^= 1; } if(k2 && !(h&1)){ d1[j] ^= 1; } if(d1[j] != d2[j]){ if(h&1) k1 ^= 1; else k2 ^= 1; ans.push_back(j); } for(int i = 0; i < v[j].size(); i++){ int e = v[j][i]; if(e != f){ dfs(e, j, k1, k2, h + 1); } } } int main(){ // freopen("in.txt", "r", stdin); int n, a, b; scanf("%d", &n); for(int i = 0; i < n - 1; i++){ scanf("%d%d", &a, &b); v[a].push_back(b); v[b].push_back(a); } for(int i = 1; i <= n; i++) scanf("%d", d1+i); for(int i = 1; i <= n; i++) scanf("%d", d2+i); dfs(1, -1, 0, 0, 0); printf("%d\n", ans.size()); for(int i = 0; i < ans.size(); i++) printf("%d\n", ans[i]); return 0; }