Submission #734876

#	Time	Username	Problem	Language	Result	Execution time	Memory
734876		Magikarp4000	Quality Of Living (IOI10_quality)	C++17	60 / 100	5058 ms	34688 KiB

quality

#include "quality.h"
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
#define OPTM ios_base::sync_with_stdio(0); cin.tie(0);
#define INF int(1e9+7)
#define ln '\n' 
#define ll long long
#define ull unsigned long long
#define ui unsigned int
#define us unsigned short
#define FOR(i,s,n) for (int i = s; i < n; i++)
#define FORR(i,n,s) for (int i = n; i > s; i--)
#define FORX(u, arr) for (auto u : arr)
#define PB push_back
#define in(v,x) (v.find(x) != v.end())
#define F first
#define S second
#define PII pair<int, int>
#define PLL pair<ll, ll>
#define UM unordered_map
#define US unordered_set
#define PQ priority_queue
#define ALL(v) v.begin(), v.end()
const ll LLINF = 1e18+1;

const int MAXN = 3e3+1;
int g[MAXN][MAXN];
ordered_set s;
int r,c;

void print() {
	FOR(k,0,r) {
		FOR(l,0,c) {
			if (s.find(g[k][l]) != s.end()) cout << 1;
			else cout << 0;
		}
		cout << ln;
	}
	cout << ln;
}

int rectangle(int R, int C, int h, int w, int G[3001][3001]) {
	r = R; c = C;
	FOR(i,0,r) FOR(j,0,c) g[i][j] = G[i][j];
	int res = INF;
	FOR(i,0,h) {
		FOR(j,0,w) s.insert(g[i][j]);
	}
	int idx = h*w/2;
	res = *s.find_by_order(idx);
	FOR(i,0,r-h+1) {
		if (i%2 == 0) {
			FOR(j,w,c) {
				//print();
				FOR(k,i,i+h) {
					s.insert(g[k][j]);
					s.erase(g[k][j-w]);
				}
				res = min(res, *s.find_by_order(idx));
			}
			if (i+h < r) {
				//print();
				FOR(j,c-w,c) {
					s.insert(g[i+h][j]);
					s.erase(g[i][j]);
				}
				res = min(res, *s.find_by_order(idx));
			}
		}
		else {
			FORR(j,c-1,w-1) {
				//print();
				FOR(k,i,i+h) {
					s.insert(g[k][j-w]);
					s.erase(g[k][j]);
				}
				res = min(res, *s.find_by_order(idx));
			}
			if (i+h < r) {
				//print();
				FOR(j,0,w) {
					s.insert(g[i+h][j]);
					s.erase(g[i][j]);
				}
				res = min(res, *s.find_by_order(idx));
			}
		}
	}
	return res;
}

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