Submission #601202

#	Time	Username	Problem	Language	Result	Execution time	Memory
601202		MohamedFaresNebili	Rectangles (IOI19_rect)	C++14	10 / 100	426 ms	460 KiB

rect

#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[400004];
            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]);
            }
            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));
            }
            ll count_rectangles(vector<vector<int>> A) {
                int N = A.size(), M = A[0].size();
              	if(N < 3) return 0;
                ll res = 0;
                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]);
                    }
                }
                return res;
            }

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