題意：求以x為根的子樹中的最小權值

思路：dfs序，把樹轉化成一個序列，類似LCA中的轉化，記錄1棵子樹進去的時間戳和出去的時間戳。線段樹求區間最小值

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<stdlib.h>
#include<math.h>
#include<vector>
#include<list>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<algorithm>
#include<numeric>
#include<functional>
using namespace std;
#define lson(x) 2*x
#define rson(x) 2*x+1
typedef long long ll;
typedef pair<int,int> pii;
const int maxn = 100005;
int in[maxn],out[maxn];
struct Edge
{
	int to,next;
}edge[maxn];
int head[maxn],tot,t;
int a[maxn],num[maxn];
void init()
{
	memset(head,-1,sizeof head);
	tot = 0;
	t = 0;
}

void add(int x,int y)
{
	edge[tot].to = y;
	edge[tot].next = head[x];
	head[x] = tot++;
}

void dfs(int x)
{
	in[x] = ++t;
	num[t] = a[x];
	for(int i = head[x]; i != -1; i = edge[i].next)
	{
		int y = edge[i].to;
		dfs(y);
	}
	out[x] = t;	
}

struct data
{
	int l,r,num,laz;
}node[maxn*4];

void pushup(int cnt)
{
	node[cnt].num = min(node[lson(cnt)].num,node[rson(cnt)].num);
}

void build(int x,int y, int cnt)
{
	node[cnt].l = x;
	node[cnt].r = y;
	if(x == y)
	{
		node[cnt].num = num[x];
		return;
	}
	int mid = (x+y) / 2;
	build(x,mid,lson(cnt));
	build(mid+1,y,rson(cnt));
	pushup(cnt);
}

void up(int x,int y,int cnt,int val)
{
	if(x == node[cnt].l && y == node[cnt].r)
	{
		node[cnt].num = val;
		return;
	}
	int mid = (node[cnt].l + node[cnt].r) / 2;
	if(y <= mid)
		up(x,y,lson(cnt),val);
	else if(x >= mid+1)
		up(x,y,rson(cnt),val);
	else
	{
		up(x,mid,lson(cnt),val);
		up(mid+1,y,rson(cnt),val);
	}	
	pushup(cnt);
}

int fid(int x,int y,int cnt)
{
	if(x == node[cnt].l && y == node[cnt].r)
	{
		return node[cnt].num;
	}
	int mid = (node[cnt].l + node[cnt].r) / 2;
	if(y <= mid)
		return fid(x,y,lson(cnt));
	else if(x >= mid+1)
		return fid(x,y,rson(cnt));
	else
		return min( fid(x,mid,lson(cnt)) , fid(mid+1,y,rson(cnt)));
}

int main(void)
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		init();
		for(int i = 1; i <= n; i++)
			scanf("%d",&a[i]);
		for(int i = 2; i <= n; i++)
		{
			int k; 
			scanf("%d",&k);
			add(k,i);
		}
		dfs(1);
		build(1,n,1);
		int q;
		scanf("%d",&q);
		while(q--)
		{
			int op,x,u;
			scanf("%d",&op);
			if(op == 1)
			{
				scanf("%d%d",&x,&u);
				up(in[x],in[x],1,u);
			}
			else
			{
				scanf("%d",&x);
				printf("%d\n",fid(in[x],out[x],1));
			}
		}
	}
	return 0;
}