Submission #1237195

#	Time	Username	Problem	Language	Result	Execution time	Memory
1237195		Muhammad_Aneeq	Rectangles (IOI19_rect)	C++20	59 / 100	1850 ms	1083328 KiB

rect

#pragma GCC optimize("O2")
#include "rect.h"
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pii pair<int,int>
int const N=2501,LG=12;
struct spmx //sparse table for max
{
	int sp[N][LG]={};
	spmx(vector<int>vals)
	{
		int n=vals.size();
		for (int i=0;i<n;i++)
			sp[i][0]=vals[i];
		for (int j=1;(1<<j)<=n;j++)
			for (int i=0;i+(1<<j)-1<n;i++)
				sp[i][j]=max(sp[i][j-1],sp[i+(1<<(j-1))][j-1]);
	}
	inline int get(int l,int r)
	{
		if (l>r)
			return 1e9;
		int lg=log2(r-l+1);
		return max(sp[l][lg],sp[r+1-(1<<lg)][lg]);
	}
};
struct spmn //sparse table for min
{
	int sp[N][LG]={};
	spmn(vector<int>vals)
	{
		int n=vals.size();
		for (int i=0;i<n;i++)
			sp[i][0]=vals[i];
		for (int j=1;(1<<j)<=n;j++)
			for (int i=0;i+(1<<j)-1<n;i++)
				sp[i][j]=min(sp[i][j-1],sp[i+(1<<(j-1))][j-1]);
	}
	 inline int get(int l,int r)
	{
		if (l>r)
			return 0;
		int lg=log2(r-l+1);
		return min(sp[l][lg],sp[r+1-(1<<lg)][lg]);
	}
};
int l[N][N],r[N][N],u[N][N],d[N][N]={};
int n,m;
vector<spmn>mr,md;
vector<spmx>ml,mu;
inline bool check(int u,int d,int l,int r)
{
	if (u==0||d==n-1||l==0||r==m-1)
		return 0;
	if (md[u-1].get(l,r)<d)
		return 0;
	if (mu[d+1].get(l,r)>u)
		return 0;
	if (ml[r+1].get(u,d)>l)
		return 0;
	if (mr[l-1].get(u,d)<r)
		return 0;
	return 1;
}
long long count_rectangles(vector<vector<int>>a) 
{
	n=a.size(),m=a[0].size();
	vector<pair<pii,pii>>ans;
	for (int i=0;i<n;i++)
	{
		deque<pair<int,int>>vl;
		vl.push_back({-1,1e9});
		for (int j=0;j<m;j++)
		{
			while (vl.size()&&vl.back().second<a[i][j])
				vl.pop_back();
			l[i][j]=vl.back().first+1;
			vl.push_back({j,a[i][j]});
		}
		vl={};
		vl.push_back({m,1e9});
		for (int j=m-1;j>=0;j--)
		{
			while (vl.size()&&vl.back().second<a[i][j])
				vl.pop_back();
			r[i][j]=vl.back().first-1;
			vl.push_back({j,a[i][j]});
		}
	}
	for (int i=0;i<m;i++)
	{
		deque<pair<int,int>>vl;
		vl.push_back({-1,1e9});
		for (int j=0;j<n;j++)
		{
			while (vl.size()&&vl.back().second<a[j][i])
				vl.pop_back();
			u[j][i]=vl.back().first+1;
			vl.push_back({j,a[j][i]});
		}
		vl={};
		vl.push_back({n,1e9});
		for (int j=n-1;j>=0;j--)
		{
			while (vl.size()&&vl.back().second<a[j][i])
				vl.pop_back();
			d[j][i]=vl.back().first-1;
			vl.push_back({j,a[j][i]});
		}
	}
	for (int i=0;i<n;i++)
	{
		vector<int>u1,d1;
		for (int j=0;j<m;j++)
		{
			u1.push_back(u[i][j]);
			d1.push_back(d[i][j]);
		}
		mu.push_back(spmx(u1));
		md.push_back(spmn(d1));
	}
	for (int i=0;i<m;i++)
	{
		vector<int>l1,r1;
		for (int j=0;j<n;j++)
		{
			l1.push_back(l[j][i]);
			r1.push_back(r[j][i]);
		}
		ml.push_back(spmx(l1));
		mr.push_back(spmn(r1));
	}
	for (int i=0;i<n;i++)
	{
		deque<pair<int,int>>vl;
		vl.push_back({-1,1e9});
		for (int j=0;j<m;j++)
		{
			while (vl.size()&&vl.back().second<=a[i][j])
				vl.pop_back();
			l[i][j]=vl.back().first+1;
			vl.push_back({j,a[i][j]});
		}
		vl={};
		vl.push_back({m,1e9});
		for (int j=m-1;j>=0;j--)
		{
			while (vl.size()&&vl.back().second<=a[i][j])
				vl.pop_back();
			r[i][j]=vl.back().first-1;
			vl.push_back({j,a[i][j]});
		}
	}
	for (int i=0;i<m;i++)
	{
		deque<pair<int,int>>vl;
		vl.push_back({-1,1e9});
		for (int j=0;j<n;j++)
		{
			while (vl.size()&&vl.back().second<=a[j][i])
				vl.pop_back();
			u[j][i]=vl.back().first+1;
			vl.push_back({j,a[j][i]});
		}
		vl={};
		vl.push_back({n,1e9});
		for (int j=n-1;j>=0;j--)
		{
			while (vl.size()&&vl.back().second<=a[j][i])
				vl.pop_back();
			d[j][i]=vl.back().first-1;
			vl.push_back({j,a[j][i]});
		}
	}
	for (int i=1;i<n-1;i++)
	{
		for (int j=1;j<m-1;j++)
		{
			if (check(u[i][j],d[i][j],l[i][j],r[i][j]))
			{
				ans.push_back({{u[i][j],d[i][j]},{l[i][j],r[i][j]}});
			}
		}
	}
	sort(begin(ans),end(ans));
	int cnt=0;
	for (int i=0;i<ans.size();i++)
	{
		int j=i;
		while (j<ans.size()&&ans[i]==ans[j])
			j++;
		cnt++;
		i=j-1;
	}
	return cnt;
}

• Subtask 1: 8 / 8
• Subtask 2: 7 / 7
• Subtask 3: 12 / 12
• Subtask 4: 22 / 22
• Subtask 5: 10 / 10
• Subtask 6: 0 / 13
• Subtask 7: 0 / 28