제출 #203461

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
203461		stefdasca	Rectangles (IOI19_rect)	C++14	13 / 100	601 ms	86264 KiB

rect

#include "rect.h"
#include<bits/stdc++.h>
using namespace std;
long long count_rectangles(vector<vector<int> > a)
{
    int n = a.size();
    int m = a[0].size();
    vector<vector<int> >viz(n, vector<int>(m));
    for(int i = 0; i < n; ++i)
        for(int j = 0; j < m; ++j)
            viz[i][j] = 0;
    int max_val = 0;
    int min_val = (1<<30);
	for(int i = 0; i < n; ++i)
        for(int j = 0; j < m; ++j)
        {
            max_val = max(max_val, a[i][j]);
            min_val = min(min_val, a[i][j]);
        }
    if(max_val == min_val)
        return 0;
    long long ans = 0;
    if(max_val == 1)
    {
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < m; ++j)
            {
                if(a[i][j] == 0 && !viz[i][j])
                {
                    int minix = i, maxix = i, miniy = j, maxiy = j;
                    int cnt = 1;
                    viz[i][j] = 1;
                    deque<pair<int, int> >d;
                    d.push_back({i, j});
                    while(!d.empty())
                    {
                        pair<int, int> nod = d[0];
                        d.pop_front();
                        for(int ox = -1; ox <= 1; ++ox)
                            for(int oy = -1; oy <= 1; ++oy)
                            {
                                if(ox && oy)
                                    continue;
                                if(!ox && !oy)
                                    continue;
                                int new_x = nod.first + ox;
                                int new_y = nod.second + oy;
                                if(new_x >= 0 && new_y >= 0 && new_x < n && new_y < m)
                                {
                                    if(a[new_x][new_y] == 0 && !viz[new_x][new_y])
                                    {
                                        viz[new_x][new_y] = 1;
                                        d.push_back({new_x, new_y});
                                        minix = min(minix, new_x);
                                        maxix = max(maxix, new_x);
                                        miniy = min(miniy, new_y);
                                        maxiy = max(maxiy, new_y);
                                        ++cnt;
                                    }
                                }
                            }
                    }
                    if(minix && miniy && maxix < n-1 && maxiy < m-1)
                    {
                        if(cnt == (maxix - minix + 1) * (maxiy - miniy + 1))
                            ++ans;
                    }
                }
            }
    }
	return ans;
}

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