Submission #425362

#	Time	Username	Problem	Language	Result	Execution time	Memory
425362		dreezy	Quality Of Living (IOI10_quality)	C++17	60 / 100	5045 ms	19988 KiB

quality

#include <bits/stdc++.h>
//#include "quality.h"
using namespace std;

#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;

template <class T> using set_ = tree<T, null_type, less<T>,
	rb_tree_tag, tree_order_statistics_node_update>;



const int maxn = 9e6 + 5;

/*
int getmedian(){
	int cnt = 0;
	int median = -1;
	for(int a : active){
		if(cnt == active.size() / 2){
			median = a;
			break;
		}
		cnt++;
	}
	return median;
}
*/
set_<int> active;
int rectangle(int R, int C, int H, int W, int Q[3001][3001]) {
	
    active.clear();
	for(int top = 0; top < H; top ++){
		for(int left = 0; left < W; left++ ){
			active.insert(Q[top][left]);
		}
	}
	int mid = (H * W ) /2 ;
	//cout << mid<<endl;
	
	int ans = *active.find_by_order(mid);
	//cout << active.qry(mid)<<endl;
	for(int top = 0; top <= R - H; top++){
		
		//cout << top <<":\n";
		//if top % 2 === 0, we go from left to right
		
		if(top % 2 == 0){
			
			if(top > 0){
				for(int left = 0; left < W; left++ ){
					active.erase(Q[top - 1][left]);
					active.insert(Q[top + H - 1][left]);
				}
				ans = min(ans, *active.find_by_order(mid));
			}
			for(int left = 1; left <=C - W;left++){
			//iterate left boundary of rectangle
				//cout << left << "->";
				for(int t = top; t < top + H; t++){
					active.erase(Q[t][left - 1]);
					active.insert(Q[t][left + W - 1]);
				}
				ans = min(ans, *active.find_by_order(mid));
			}
		}
		else{
			if(top > 0){
				for(int right =  C - 1; right >= C - W; right-- ){
					active.erase(Q[top - 1][right]);
					active.insert(Q[top + H - 1][right]);
				}
				ans = min(ans, *active.find_by_order(mid));
			}
			//we go from right to left
			for(int right = C  - 2; right >=W - 1; right--){
				//cout << right <<"->";
			//iterate left boundary of rectangle
				for(int t = top; t < top + H; t++){
					active.erase(Q[t][right + 1]);
					active.insert(Q[t][right - W +1]);
				}
				ans = min(ans, *active.find_by_order(mid));
			}
		}
	
	//cout << endl;
	}
	
	return ans;

}

/***********/

• Subtask 1: 20 / 20
• Subtask 2: 20 / 20
• Subtask 3: 20 / 20
• Subtask 4: 0 / 20
• Subtask 5: 0 / 20