제출 #48552

#제출 시각아이디문제언어결과실행 시간메모리
48552aleksamiBirthday gift (IZhO18_treearray)C++14
100 / 100
1326 ms262144 KiB
#include <bits/stdc++.h>

using namespace std;
#define MAXN 200005
#define LOG 20
vector <int> g[MAXN];
int a[MAXN];
int in[MAXN];
int out[MAXN];
int rmq[MAXN][LOG];

int T=0;

void dfs(int v,int p)
{
	in[v]=++T;
	rmq[v][0]=p;
	for(auto x:g[v])
	{
		if(x==p)continue;
		dfs(x,v);
	}
	out[v]=T;
}

int insubtree(int u,int v)
{
	return (in[u]<=in[v]&&out[u]>=in[v]);
}

int lca(int u,int v)
{
	if(insubtree(u,v))return u;
	if(insubtree(v,u))return v;
	for(int j = LOG-1; j >= 0; j--)
	{
		if(!insubtree(rmq[u][j],v))u=rmq[u][j];
	}
	return rmq[u][0];
}

set <int> st[MAXN];
set <int> stz[MAXN];

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	int n,m,q;
	cin >> n >> m >> q;
	for(int i = 0; i < n-1; i++)
	{
		int u,v;
		cin >> u >> v;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	dfs(1,0);
	rmq[1][0]=1;
	for(int d = 1; d < LOG; d++)
	{
		for(int i = 1; i <= n; i++)rmq[i][d]=rmq[rmq[i][d-1]][d-1];
	}
	for(int i = 1; i <= m; i++)cin >> a[i];
	for(int i = 1; i <= m-1; i++)
	{
		st[lca(a[i],a[i+1])].insert(i);
	}
	for(int i = 1; i <= m; i++)
	{
		stz[a[i]].insert(i);
	}
	while(q--)
	{
		int t;
		cin >> t;
		if(t==1)
		{
			int idx,val;
			cin >> idx >> val;
			if(idx!=m)
			{
				st[lca(a[idx],a[idx+1])].erase(idx);
				st[lca(val,a[idx+1])].insert(idx);
			}
			if(idx != 1)
			{
				st[lca(a[idx-1],a[idx])].erase(idx-1);
				st[lca(a[idx-1],val)].insert(idx-1);
			}
			stz[a[idx]].erase(idx);
			stz[val].insert(idx);
			a[idx]=val;
		}
		else
		{
			int l,r,v;
			cin >> l >> r >> v;
			auto it = st[v].lower_bound(l);
			auto itt = stz[v].lower_bound(l);
			if(itt!=stz[v].end() && *itt >= l && *itt <= r)
			{
				cout << *itt << " " << *itt << "\n";
				continue;
			}
			if(it==st[v].end())
			{
				cout << "-1 -1\n";
				continue;
			}
			if(*it <= r && *it >= l && (*it+1)<=r)
			{
				cout << *it << " " << *it+1 << "\n";
			}
			else
			{
				cout << "-1 -1\n";
			}
		}
	}
	return 0;
}
/*
5 4 4
1 2
3 1
3 4
5 3
4 5 2 3
2 1 3 1
1 3 5
2 3 4 5
2 1 3 1

*/
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