Submission #297677

#	Time	Username	Problem	Language	Result	Execution time	Memory
297677		shayan_p	Rectangles (IOI19_rect)	C++17	50 / 100	5017 ms	172408 KiB

rect

#include<bits/stdc++.h>
#include "rect.h"
     
#define F first
#define S second
#define PB push_back
#define sz(s) (int(s.size()))
#define bit(n, k) (((n)>>(k))&1)
     
using namespace std;
     
typedef long long ll;
typedef pair<int, int> pii;
     
const int maxn = 2510;
     
vector<vector<int>> a;
int aftx[maxn][maxn], befx[maxn][maxn], afty[maxn][maxn], befy[maxn][maxn];
int n, m;
     
bool ok(int LX, int RX, int LY, int RY, int i, int j){
    if(LX <= 0 || LY <= 0 || RX >= n-1 || RY >= m-1 || LX > RX || LY > RY)
    	return 0;    
    bool ok = 1;
    for(int x = LX; x <= RX; x++){
    	for(int y = LY; y <= RY; y++){
    	    ok&= a[i][j] >= a[x][y];
    	    if((pii){x, y} > (pii){i, j})
    		ok&= a[i][j] > a[x][y];
    	    ok&= befx[x][y] >= LY - 1;
    	    ok&= befy[x][y] >= LX - 1;
    	    ok&= aftx[x][y] <= RY + 1;
    	    ok&= afty[x][y] <= RX + 1;
	    if(!ok)
		return 0;
    	}
    }
    return 1;
}
     
ll count_rectangles(vector< vector<int> > a){
    ::a = a;
    n = sz(a);
    m = sz(a[0]);
    for(int i = 0; i < n; i++){
    	for(int j = 0; j < m; j++){
    	    befx[i][j] = j-1;
    	    while(befx[i][j] >= 0 && a[i][befx[i][j]] <= a[i][j])
    		befx[i][j] = befx[i][ befx[i][j] ];
    	}
    	for(int j = m-1; j >= 0; j--){
    	    aftx[i][j] = j+1;
    	    while(aftx[i][j] < m && a[i][aftx[i][j]] <= a[i][j])
    		aftx[i][j] = aftx[i][ aftx[i][j] ];
    	}
    }
    for(int j = 0; j < m; j++){
    	for(int i = 0; i < n; i++){
    	    befy[i][j] = i-1;
    	    while(befy[i][j] >= 0 && a[befy[i][j]][j] <= a[i][j])
    		befy[i][j] = befy[ befy[i][j] ][j];
    	}
    	for(int i = n-1; i >= 0; i--){
    	    afty[i][j] = i+1;
    	    while(afty[i][j] < n && a[afty[i][j]][j] <= a[i][j])
    		afty[i][j] = afty[ afty[i][j] ][j];
    	}	
    }
        
    ll ans = 0;
    for(int i = 0; i < n; i++){
    	for(int j = 0; j < m; j++){
    	    ans+= ok(befy[i][j] + 1, afty[i][j] - 1, befx[i][j] + 1, aftx[i][j] - 1, i, j);
    	}
    }
    return ans;
}

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