题意：给定 n 个盒子，然后告诉你每个盒子在哪个盒子里，数值越大，盒子越大，给定你初态，和末态，问你最少要几步能完成，只有两种操作，一种是把一个盒子连同里面的小盒子放到一个空盒子里，另一种是把一个堆盒子里的最外面的那个盒子拿出来。
析：首先，先遍历一次，如果初态和不一样，那么初态后面的要全部拿出来，然后再遍历一次，然后如果发现不一样，然后要看把末态的父结点是不是孤立的，如果不是，也要全部拿出。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
//#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std; typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e15;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 100;
const int mod = 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
} int fa[maxn];
int pa[maxn];
int ans; void dfs(int x){
if(pa[x] == 0) return ;
dfs(pa[x]); pa[x] = 0;
++ans;
} int main(){
while(scanf("%d", &n) == 1){
for(int i = 1; i <= n; ++i) scanf("%d", pa+i);
for(int i = 1; i <= n; ++i) scanf("%d", fa+i);
ans = 0;
for(int i = 1; i <= n; ++i) // remove
if(pa[i] != fa[i]) dfs(i);
for(int i = 1; i <= n; ++i){ // unit
if(pa[i] == fa[i]) continue;
++ans; dfs(fa[i]);
}
printf("%d\n", ans);
}
return 0;
}