제출 #1080442

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1080442		speedcode	Rectangles (IOI19_rect)	C++17	0 / 100	73 ms	51028 KiB

rect

#include <bits/stdc++.h>
using namespace std;

long long count_rectangles(std::vector<std::vector<int>> a)
{
    long long n = a.size();
    long long m = a[0].size();

    vector<vector<int>> explored(n);
    for(int i = 0; i < n; i++) explored[i].resize(m);

    for(int i = 0; i < n; i++)
    for(int j = 0; j < m; j++)
        explored[i][j] = 0;

    vector<vector<int>> prefix(n+1);
    for(int i = 0; i < n+1; i++) prefix[i].resize(m+1);

    prefix[0][0] = 0;
    for(int i = 0; i < n+1; i++) prefix[i][0] = 0;
    for(int i = 0; i < m+1; i++) prefix[0][i] = 0;

    for(int i = 0; i < n; i++){
        for(int j = 0; j < m; j++){
            prefix[i+1][j+1] = a[i][j];
            prefix[i+1][j+1] += prefix[i][j+1];
            prefix[i+1][j+1] += prefix[i+1][j];
            prefix[i+1][j+1] -= prefix[i][j];
        }
    }


    long long res = 0;

    for(long long i = 1; i < n-1; i++){
        for(long long j = 1; j < m-1; j++){
            if(explored[i][j]) continue;
            explored[i][j] = 1;
            if(a[i][j] > 0){
                continue;
            }

            long long endX = i;
            long long endY = j;

            while(endX < n-2 && a[endX+1][endY] == 0){
                endX++;
                explored[endX][j] = 1;
            }

            while(endY < m-2 && a[endX][endY+1] == 0){
                endY++;
                for(int u = i; u <= endX; u++){
                    explored[u][endY] = 1;
                }
            }



            if(prefix[endX+1][endY+1] + prefix[i-1+1][j-1+1] - prefix[i-1+1][endY+1]-prefix[endX+1][j-1+1] != 0) continue;

            
            long long nbOnes = 0;

            nbOnes += prefix[endX+2][endY+2] + prefix[i-1][j-1] - prefix[i-1][endY+2]-prefix[endX+2][j-1];
            nbOnes -= prefix[endX+1][endY+1] + prefix[i-1+1][j-1+1] - prefix[i-1+1][endY+1]-prefix[endX+1][j-1+1];
            nbOnes -= a[i-1][j-1];
            nbOnes -= a[endX+1][j-1];
            nbOnes -= a[i-1][endY+1];
            nbOnes -= a[endX+1][endY+1];

            if(nbOnes == 2*(endX-i+1) + 2*(endY-j+1)){
                 res++;
            }
        }
    }

    return res;
}

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