Submission #1030896

#	Time	Username	Problem	Language	Result	Execution time	Memory
1030896		Zicrus	Rectangles (IOI19_rect)	C++17	25 / 100	5096 ms	176464 KiB

rect

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

typedef long long ll;

ll n, m;
ll pow2N, pow2M;
vector<vector<ll>> segN, segM;

ll segMaxN(ll low, ll high, ll m) {
    low += pow2N; high += pow2N;
    ll mx = 0;
    while (low <= high) {
        if (low & 1) mx = max(mx, segN[m][low++]);
        if (!(high & 1)) mx = max(mx, segN[m][high--]);
        low /= 2; high /= 2;
    }
    return mx;
}

ll segMaxM(ll low, ll high, ll n) {
    low += pow2M; high += pow2M;
    ll mx = 0;
    while (low <= high) {
        if (low & 1) mx = max(mx, segM[n][low++]);
        if (!(high & 1)) mx = max(mx, segM[n][high--]);
        low /= 2; high /= 2;
    }
    return mx;
}

ll count_rectangles(vector<vector<int>> a) {
    if (a.size() <= 2) return 0;
    n = a.size(); m = a[0].size();
    pow2N = 1ll << (ll)ceil(log2(n));
    pow2M = 1ll << (ll)ceil(log2(m));
    vector<int> am = a[1];
    segN = vector<vector<ll>>(m, vector<ll>(2*pow2N));
    segM = vector<vector<ll>>(n, vector<ll>(2*pow2M));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            segN[j][pow2N+i] = a[i][j];
            segM[i][pow2M+j] = a[i][j];
        }
    }
    for (int meta = 0; meta < n; meta++) {
        for (int i = pow2M-1; i > 0; i--) {
            segM[meta][i] = max(segM[meta][2*i], segM[meta][2*i+1]);
        }
    }
    for (int meta = 0; meta < m; meta++) {
        for (int i = pow2N-1; i > 0; i--) {
            segN[meta][i] = max(segN[meta][2*i], segN[meta][2*i+1]);
        }
    }

    ll res = 0;
    for (int lN = 1; lN <= n; lN++) {
        for (int lM = 1; lM <= m; lM++) {
            for (int i = 1; i+lN-1 < n-1; i++) {
                for (int j = 1; j+lM-1 < m-1; j++) {
                    
                    bool poss = 1;
                    for (int meta = i; meta < i+lN && poss; meta++) {
                        ll border = min(a[meta][j-1], a[meta][j+lM]);
                        if (border <= segMaxM(j, j+lM-1, meta)) {
                            poss = 0;
                        }
                    }
                    for (int meta = j; meta < j+lM && poss; meta++) {
                        ll border = min(a[i-1][meta], a[i+lN][meta]);
                        if (border <= segMaxN(i, i+lN-1, meta)) {
                            poss = 0;
                        }
                    }
                    res += poss;
                    if (poss) j += lM;
                }
            }
        }
    }
    return res;
}

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