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,

//qscqesze #include #include #include #include #include #include #include <set> #include #include #include #include #include #include #include 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("\n"); } */ //************************************************************************************** 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) { int v=e[x][i]; dfs(v,d,c); } } else { ans.push_back(x); for(int i=0;i) { int v=e[x][i]; dfs(v,d,!c); } } } int main() { int n; cin>>n; for(int i=0;i1;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<endl; for(int i=0;i) cout<endl; }

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