贪心：只要统计有多少个子节点与它的父节点不一样

时间复杂度：\({\Large \Theta(n)}\)

#include <bits/stdc++.h>

using namespace std;

const int N = 200009;

int n;
int a[N], f[N], p[N];

int main(void) {
    cin >> n;
    for(int i = 2; i <= n; i ++ ) cin >> p[i];
    for(int i = 1; i <= n; i ++ ) cin >> a[i];
    
    int ans = n;
    for(int i = 2; i <= n; i ++ )
        if(a[i] == a[p[i]]) ans --;
        
    cout << ans << endl;
    return 0;
}