Submission #297977

#	Time	Username	Problem	Language	Result	Execution time	Memory
297977		erd1	Rectangles (IOI19_rect)	C++14	59 / 100	4554 ms	1048580 KiB

rect

#include "rect.h"
#include<bits/stdc++.h>
using namespace std;
#define ff first
#define ss second
#define pb push_back
#define all(x) (x).begin(), (x).end()
typedef int64_t lld;
typedef pair<int, int> pii;

/*
#include<bits/extc++.h>
using namespace __gnu_pbds;
template<typename T, typename comp = greater<T>>
using OST = tree<T, null_type, comp, rb_tree_tag, tree_order_statistics_node_update>;
*/

template<typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> p){
	return out << "(" << p.ff << ", " << p.ss << ")";
}

vector<pii> tmp;
vector<vector<int>> U, D, L, R;

template<typename Cmp>
struct node{
	node *L, *R;
	int l, r, mid;
	int cmp(int a, int b){ return Cmp()(a, b)?a:b; }
	struct inode{
		inode *L, *R;
		int l, r, mid;
		int val;
		int cmp(int a, int b){ return Cmp()(a, b)?a:b; }
		inode(int ll, int rr, const vector<int>& v) : l(ll), r(rr), mid((l+r)/2) {
			if(l == r)val = v[l];
			else L = new inode(l, mid, v), R = new inode(mid+1, r, v), val = cmp(L->val, R->val);
		}
		inode(int ll, int rr, inode* n1, inode* n2) : l(ll), r(rr), mid((l+r)/2) {
			val = cmp(n1->val, n2->val);
			if(l != r)L = new inode(l, mid, n1->L, n2->L), R = new inode(mid+1, r, n1->R, n2->R);
		}
		int query(int ll, int rr){
			assert(ll <= r && rr >= l);
			if(ll <= l && rr >= r)return val;
			if(ll > mid)return R->query(ll, rr);
			if(rr <= mid)return L->query(ll, rr);
			return cmp(L->query(ll, rr), R->query(ll, rr));
		}
	} *data;
	node(int ll, int rr, int l2, int r2, const vector<vector<int>>& v) : l(ll), r(rr), mid((l+r)/2) {
		if(l != r) L = new node(l, mid, l2, r2, v), R = new node(mid+1, r, l2, r2, v), data = new inode(l2, r2, L->data, R->data);
		else data = new inode(l2, r2, v[l]);
	}
	int query(int l1, int r1, int l2, int r2){
		assert(l1 <= r && r1 >= l);
		if(l1 <= l && r1 >= r)return data->query(l2, r2);
		if(l1 > mid)return R->query(l1, r1, l2, r2);
		if(r1 <= mid)return L->query(l1, r1, l2, r2);
		return cmp(L->query(l1, r1, l2, r2), R->query(l1, r1, l2, r2));
	}
};

node<greater<int>> *d, *r;
node<less<int>> *u, *l;
set<tuple<int, int, int, int>> st;

long long count_rectangles(vector<vector<int>> a) {
	int n = a.size(), m = a[0].size();
	U = D = L = R = a;
	for(int i = 1; i < n; i++){
		tmp.resize(0); tmp.pb({INT_MAX, -1});
		for(int j = 0; j < m; j++){
			while(tmp.back().ff <= a[i][j])tmp.pop_back();
			L[i][j] = tmp.back().ss;
			tmp.pb({a[i][j], j});
		}
		tmp.resize(0); tmp.pb({INT_MAX, m});
		for(int j = m-1; j > 0; j--){
			while(tmp.back().ff <= a[i][j])tmp.pop_back();
			R[i][j] = tmp.back().ss;
			tmp.pb({a[i][j], j});
		}
	}
	for(int j = 1; j < m; j++){
		tmp.resize(0); tmp.pb({INT_MAX, -1});
		for(int i = 0; i < n; i++){
			while(tmp.back().ff <= a[i][j])tmp.pop_back();
			U[i][j] = tmp.back().ss;
			tmp.pb({a[i][j], i});
		}
		tmp.resize(0); tmp.pb({INT_MAX, n});
		for(int i = n-1; i > 0; i--){
			while(tmp.back().ff <= a[i][j])tmp.pop_back();
			D[i][j] = tmp.back().ss;
			tmp.pb({a[i][j], i});
		}
	}
	u = new node<less<int>>(0, n-1, 0, m-1, U), l = new node<less<int>>(0, n-1, 0, m-1, L);
	d = new node<greater<int>>(0, n-1, 0, m-1, D), r = new node<greater<int>>(0, n-1, 0, m-1, R);
	for(int i = 1; i <= n-2; i++)
		for(int j = 1; j <= m-2; j++){
			if(U[i][j] < 0 || D[i][j] >= n || L[i][j] < 0 || R[i][j] >= m)goto next;
			//cout << i << " " << j << " " << U[i][j] << D[i][j] << L[i][j] << R[i][j] << " " << u->query(U[i][j]+1, D[i][j]-1, L[i][j]+1, R[i][j]-1) << endl;
			if(u->query(U[i][j]+1, D[i][j]-1, L[i][j]+1, R[i][j]-1) != U[i][j])goto next;
			if(l->query(U[i][j]+1, D[i][j]-1, L[i][j]+1, R[i][j]-1) != L[i][j])goto next;
			if(d->query(U[i][j]+1, D[i][j]-1, L[i][j]+1, R[i][j]-1) != D[i][j])goto next;
			if(r->query(U[i][j]+1, D[i][j]-1, L[i][j]+1, R[i][j]-1) != R[i][j])goto next;
			st.insert({U[i][j], L[i][j], D[i][j], R[i][j]});
			next:;
		}
	return st.size();
}

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