제출 #601248

#제출 시각아이디문제언어결과실행 시간메모리
601248MohamedFaresNebiliRectangles (IOI19_rect)C++14
18 / 100
5046 ms51592 KiB
#include <bits/stdc++.h>
#include "rect.h"
/// #pragma GCC optimize ("Ofast")
/// #pragma GCC target ("avx2")
/// #pragma GCC optimize("unroll-loops")

            using namespace std;

            using ll = long long;
            using ii = pair<ll, ll>;
            using vi = vector<int>;

            #define ff first
            #define ss second
            #define pb push_back
            #define all(x) (x).begin(), (x).end()
            #define lb lower_bound

            const int oo = 1000 * 1000 * 1000 + 7;

            int ST[40004], SP[120][120], SG[120][120];
            void update(int v, int l, int r, int p, int val) {
                if(l == r) {
                    ST[v] = val;
                    return;
                }
                int md = (l + r) / 2;
                if(p <= md) update(v * 2, l, md, p, val);
                else update(v * 2 + 1, md + 1, r, p, val);
                ST[v] = max(ST[v * 2], ST[v * 2 + 1]);
            }
            void update(int c, int v, int l, int r, int p, int val) {
                if(l == r) {
                    SP[v][c] = val;
                    return;
                }
                int md = (l + r) / 2;
                if(p <= md) update(c, v * 2, l, md, p, val);
                else update(c, v * 2 + 1, md + 1, r, p, val);
                SP[v][c] = max(SP[v * 2][c], SP[v * 2 + 1][c]);
            }
            void Update(int c, int v, int l, int r, int p, int val) {
                if(l == r) {
                    SG[v][c] = val;
                    return;
                }
                int md = (l + r) / 2;
                if(p <= md) Update(c, v * 2, l, md, p, val);
                else Update(c, v * 2 + 1, md + 1, r, p, val);
                SG[v][c] = max(SG[v * 2][c], SG[v * 2 + 1][c]);
            }
            int query(int v, int l, int r, int lo, int hi) {
                if(l > hi || r < lo) return 0;
                if(l >= lo && r <= hi) return ST[v];
                return max(query(v * 2, l, (l + r) / 2, lo, hi),
                           query(v * 2 + 1, (l + r) / 2 + 1, r, lo, hi));
            }
            int query(int c, int v, int l, int r, int lo, int hi) {
                if(l > hi || r < lo) return 0;
                if(l >= lo && r <= hi) return SP[v][c];
                return max(query(c, v * 2, l, (l + r) / 2, lo, hi),
                           query(c, v * 2 + 1, (l + r) / 2 + 1, r, lo, hi));
            }
            int Query(int c, int v, int l, int r, int lo, int hi) {
                if(l > hi || r < lo) return 0;
                if(l >= lo && r <= hi) return SG[v][c];
                return max(Query(c, v * 2, l, (l + r) / 2, lo, hi),
                           Query(c, v * 2 + 1, (l + r) / 2 + 1, r, lo, hi));
            }
            ll count_rectangles(vector<vector<int>> A) {
                int N = A.size(), M = A[0].size();
              	ll res = 0;
              	if(N == 3) {
                    for(int l = 1; l < M - 1; l++) {
                        if(A[1][l] >= A[0][l] || A[1][l] >= A[2][l])
                            update(1, 0, M - 1, l, 1000000007);
                        else update(1, 0, M - 1, l, A[1][l]);
                    }
                    for(int l = 1; l < M - 1; l++) {
                        for(int i = l; i < M - 1; i++) {
                            int V = query(1, 0, M - 1, l, i);
                            res += (V < A[1][l - 1] && V < A[1][i + 1]);
                        }
                    }
              	}
              	if(N > 3) {
                    for(int l = 1; l < N - 1; l++)
                        for(int i = 1; i < M - 1; i++)
                            update(l, 1, 0, M - 1, i, A[l][i]),
                            Update(i, 1, 0, N - 1, l, A[l][i]);
                    for(int l = 1; l < N - 1; l++)
                        for(int i = 1; i < M - 1; i++)
                            for(int j = l; j < N - 1; j++)
                                for(int k = i; k < M - 1; k++) {
                                    bool ok = 1;
                                    for(int x = l; x <= j; x++) {
                                        int V = query(x, 1, 0, M - 1, i, k);
                                        ok &= (V < A[x][i - 1] && V < A[x][k + 1]);
                                    }
                                    for(int y = i; y <= k; y++) {
                                        int V = Query(y, 1, 0, N - 1, l, j);
                                        ok &= (V < A[l - 1][y] && V < A[j + 1][y]);
                                    }
                                    res += ok;
                                }
              	}
                return res;
            }
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