제출 #601202

#제출 시각아이디문제언어결과실행 시간메모리
601202MohamedFaresNebiliRectangles (IOI19_rect)C++14
10 / 100
426 ms460 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[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; }
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