原题链接
考察:DFS
思路:
  \(mod[i]\)维护\(i\)结点是否被修改,\(anc[i]\)记录\(i\)结点的父节点的父节点是谁.直接一次DFS即可同时求出两个值.

Code

#include <iostream>
#include <cstring>
#include <vector>
using namespace std;
const int N = 100010;
struct Road{
	int ne,to;
}road[N<<1];
int n,h[N],idx,goal[N],source[N],res;
int anc[N],mod[N],ans[N],cnt;
void add(int a,int b)
{
	road[idx].to = b,road[idx].ne =h[a],h[a] = idx++;
}
void dfs(int u,int fa)
{
	if(goal[u]!=source[u])
	{
		if(!mod[anc[u]]) res++,ans[++cnt] = u;
		mod[u] = 1;//被修改了 
		source[u]^=1;
	}else if(mod[anc[u]]) res++,ans[++cnt] = u;
	for(int i=h[u];~i;i=road[i].ne)
	{
		int v = road[i].to;
		if(v==fa) continue;
		anc[v] = fa;
		dfs(v,u);
	}
	mod[u] = 0;
}
int main()
{
	scanf("%d",&n);
	memset(h,-1,sizeof h); 
	for(int i=1;i<n;i++)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		add(a,b),add(b,a);
	}
	for(int i=1;i<=n;i++) scanf("%d",&source[i]);
	for(int i=1;i<=n;i++) scanf("%d",&goal[i]);
	dfs(1,0);
	printf("%d\n",res);
	for(int i=1;i<=cnt;i++) printf("%d\n",ans[i]);
	return 0;
}