이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
#define rep(a, b) for(int a = 0; a < (b); ++a)
#define st first
#define nd second
#define pb push_back
#define all(a) a.begin(), a.end()
const int LIM=1e5+7;
int a[LIM], b[LIM], dp[5007][5007], ma[5007][5007], n;
int tr[4*LIM], tr2[4*LIM], N=1;
map<int,int>mp;
void subtask2() {
	rep(i, n-1) if(a[i+1]==b[1] && a[i+1]>=a[i+2]) a[i+2]=a[i+1];
	for(int i=n-1; i; --i) if(a[i+1]==b[1] && a[i+1]>=a[i]) a[i]=a[i+1];
	int ans=0;
	rep(i, n) if(a[i+1]==b[i+1]) ++ans;
	cout << ans << '\n';
}
int cnt(int v) {
	v+=N;
	int ans=tr[v];
	while(v) {
		if(v%2==1) ans=max(ans, tr[v-1]);
		v/=2;
	}
	return ans;
}
void upd(int v, int x) {
	v+=N;
	while(v) {
		tr[v]=max(tr[v], x);
		v/=2;
	}
}
int cnt2(int l, int r) {
	l+=N; r+=N;
	int ans=max(tr2[l], tr2[r]);
	while(l/2!=r/2) {
		if(l%2==0) ans=max(ans, tr2[l+1]);
		if(r%2==1) ans=max(ans, tr2[r-1]);
		l/=2; r/=2;
	}
	return ans;
}
void subtask4() {
	while(N<=n) N*=2;
	rep(i, n) tr2[N+i+1]=a[i+1];
	for(int i=N-1; i; --i) tr2[i]=max(tr2[2*i], tr2[2*i+1]);
	rep(i, n) mp[a[i+1]]=i+1;
	rep(i, n) {
		if(!mp[b[i+1]] || cnt2(min(mp[b[i+1]], i+1), max(mp[b[i+1]], i+1))>b[i+1]) continue;
		upd(mp[b[i+1]], cnt(mp[b[i+1]])+1);
	}
	cout << cnt(n) << '\n';
}
int main() {
	ios_base::sync_with_stdio(0); cin.tie(0);
	cin >> n;
	rep(i, n) cin >> a[i+1];
	rep(i, n) cin >> b[i+1];
	bool ok2=true;
	rep(i, n-1) if(b[i+1]!=b[i+2]) ok2=false;
	if(ok2) {
		subtask2();
		return 0;
	}
	if(n>5000) {
		subtask4();
		return 0;
	}
	rep(i, n) {
		for(int j=i; j<n; ++j) {
			ma[i+1][j+1]=max(ma[i+1][j], a[j+1]);
		}
	}
	rep(i, n) rep(j, n) {
		dp[i+1][j+1]=max(dp[i+1][j], dp[i][j+1]);
		if(a[i+1]==b[j+1] && ma[min(i+1, j+1)][max(i+1, j+1)]==a[i+1]) {
			dp[i+1][j+1]=max(dp[i+1][j+1], dp[i+1][j]+1);
		}
	}
	cout << dp[n][n] << '\n';
}
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