제출 #1032394

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1032394		shmax	Rectangles (IOI19_rect)	C++17	37 / 100	3067 ms	1048576 KiB

rect

#include "rect.h"
#include <bits/stdc++.h>

//#pragma GCC optimize("O3")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("avx,avx2,bmi,bmi2")

using namespace std;
using i32 = int;
//#define int long long
#define len(x) (int)(x.size())
#define inf 1000'000'000'000'000'000LL
#define all(x) x.begin(), x.end()
#define low_bit(x) (x & (-x))

template<typename T>
using vec = vector<T>;

const int maxM = 701;

long long count_rectangles(std::vector<std::vector<i32>> a) {
    int n = len(a);
    int m = len(a[0]);
    long long ans = 0;
    vec<vec<bitset<maxM>>> b(n, vec<bitset<maxM>>(m));
    for (int i = 1; i < n - 1; i++) {
        for (int j = 1; j < m - 1; j++) {
            b[i][j].reset();
            int mx = a[i][j];
            for (int l = i; l < n - 1; l++) {
                mx = max(mx, a[l][j]);
                if (mx >= min(a[i - 1][j], a[l + 1][j])) continue;
                b[i][j][l] = true;
            }

        }
    }

    vec<vec<bitset<maxM>>> good(m, vec<bitset<maxM>>(m));

    vec<vec<vec<int>>> maxVal(n, vec<vec<int>>(m, vec<int>(m, 0)));
    for (int i = 1; i < n - 1; i++) {
        for (int l = 1; l < m - 1; l++) {
            for (int r = l; r < m - 1; r++) {
                maxVal[i][l][r] = max(maxVal[i][l][r - 1], a[i][r]);
            }
        }
    }
    for (int l = 1; l < m - 1; l++) {
        for (int r = l; r < m - 1; r++) {
            for (int i = 0; i < n; i++) {
                good[l][r][i] = maxVal[i][l][r] < min(a[i][l - 1], a[i][r + 1]);
            }
        }
    }
    vec<vec<vec<int>>> lowest(m, vec<vec<int>>(m, vec<int>(n, 0)));
    for (int l = 1; l < m - 1; l++) {
        for (int r = l; r < m - 1; r++) {
            for (int i = n - 2; i >= 1; i--) {
                if (!good[l][r][i]) continue;
                lowest[l][r][i] = lowest[l][r][i + 1] + 1;
            }
        }
    }
    for (int i = 1; i < n - 1; i++) {
        for (int j = 1; j < m - 1; j++) {
            if (a[i][j] >= a[i - 1][j]) continue;
            if (a[i][j] >= a[i][j - 1]) continue;
            for (int k = j; k < m - 1; k++) {
                b[i][j] &= b[i][k];

                ans += (b[i][j] & ((good[j][k] << (maxM - i - lowest[j][k][i])) >> (maxM - i - lowest[j][k][i]))).count();
            }
        }
    }

    return ans;
}

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