제출 #937240

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
937240		GrindMachine	Rectangles (IOI19_rect)	C++17	0 / 100	31 ms	49752 KiB

rect

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*

read some solutions long back, remember some ideas from there

*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

#include "rect.h"

long long count_rectangles(std::vector<std::vector<int> > a) {
	int n = sz(a), m = sz(a[0]);
	if(n <= 2 or m <= 2) return 0;

	vector<pii> up[n][m];
	int ptr = 0;

	bool there[m][n][n];
	memset(there,0,sizeof there);

	rep(j,m){
		stack<int> st;
		rep(i,n){
			while(!st.empty() and a[i][j] >= a[st.top()][j]){
				int k = st.top();
				st.pop();

				if(i-k >= 2){
					there[j][k][i] = 1;
					up[i][j].pb({k,ptr++});
				}
			}

			if(!st.empty()){
				int k = st.top();
				if(i-k >= 2){
					there[j][k][i] = 1;
					up[i][j].pb({k,ptr++});
				}
			}

			st.push(i);
		}
	}

	vector<int> farthest(ptr);
	map<array<int,3>,int> mp;

	rev(j,m-1,0){
		rep(i,n){
			for(auto [k,ptr] : up[i][j]){
				array<int,3> ar = {j,k,i};
				array<int,3> want = {j+1,k,i};
				if(mp.count(want)){
					farthest[ptr] = farthest[mp[want]];
				}
				else{
					farthest[ptr] = j;
				}

				mp[ar] = ptr;
			}
		}
	}

	int prev_occ[m][m], first_occ[m][m];
	memset(prev_occ,-1,sizeof prev_occ);
	memset(first_occ,-1,sizeof first_occ);

	ll ans = 0;

	rep1(i,n-1){
		stack<int> st;
		vector<pii> good;
		rep(j,m){
			while(!st.empty() and a[i][j] >= a[i][st.top()]){
				int k = st.top();
				st.pop();

				if(j-k >= 2){
					good.pb({k,j});
				}
			}

			if(!st.empty()){
				int k = st.top();
				if(j-k >= 2){
					good.pb({k,j});
				}
			}

			st.push(j);
		}

		for(auto [l,r] : good){
			if(prev_occ[l][r] != i-1){
				first_occ[l][r] = i;
			}

			prev_occ[l][r] = i;

			for(int k = first_occ[l][r]; k <= i; ++k){
				if(there[l+1][k-1][i+1]){
					array<int,3> ar = {l+1,k-1,i+1};
					int ptr = mp[ar];
					if(farthest[ptr] >= r-1){
						ans++;
					}
				}
			}
		}
	}

	return ans;
}

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