Submission #155922

#TimeUsernameProblemLanguageResultExecution timeMemory
155922qkxwsm미술 수업 (IOI13_artclass)C++14
73 / 100
81 ms9720 KiB
#include "artclass.h"
#include <bits/stdc++.h>

using namespace std;

template<class T, class U>
void ckmin(T &a, U b)
{
	if (a > b) a = b;
}
template<class T, class U>
void ckmax(T &a, U b)
{
	if (a < b) a = b;
}

#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)
#define SZ(x) ((int) ((x).size()))
#define ALL(x) (x).begin(), (x).end()

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;
typedef pair<ld, ld> pdd;
typedef vector<vector<array<int, 3> > > img;

int ans;
img grid;
pdd vals[4] = {{2881.44,62.80}, {6595.89,100.27}, {7891.44,255.36}, {2518.33,22.10}};
ld co;

pdd calc(img f)
{
	//{# of distinct colors, average blackness}
	pdd res = {0, 0};
	set<array<int, 3> > s;
	FOR(i, 0, 100)
	{
		FOR(j, 0, 100)
		{
			array<int, 3> cur = f[i][j];
			// FOR(k, 0, 3) cur[i] /= 10;
			s.insert(cur);
		}
	}
	FOR(i, 1, 100)
	{
		FOR(j, 1, 100)
		{
			FOR(dx, -1, 1)
			{
				FOR(dy, -1, 1)
				{
					FOR(k, 0, 3)
					{
						res.se += abs(f[i][j][k] - f[i + dx][j + dy][k]);
					}
				}
			}
		}
	}
	res.fi = SZ(s);
	res.se /= 10000;
	return res;
}
ld dist(pdd a, pdd b)
{
	return 0.0 * (a.fi - b.fi) * (a.fi - b.fi) + co * (a.se - b.se) * (a.se - b.se);
}

int style(int H, int W, int R[500][500], int G[500][500], int B[500][500])
{
	//calculate some quantity of
	grid.resize(H);
	FOR(i, 0, H)
	{
		grid[i].resize(W);
		FOR(j, 0, W)
		{
			grid[i][j][0] = R[i][j];
			grid[i][j][1] = G[i][j];
			grid[i][j][2] = B[i][j];
		}
	}
	ld dl = max(max(vals[0].fi, vals[1].fi), max(vals[2].fi, vals[3].fi)) - min(min(vals[0].fi, vals[1].fi), min(vals[2].fi, vals[3].fi));
	ld dr = max(max(vals[0].se, vals[1].se), max(vals[2].se, vals[3].se)) - min(min(vals[0].se, vals[1].se), min(vals[2].se, vals[3].se));
	co = (dl / dr) * (dl / dr);
	auto cur = calc(grid);
	ans = 0;
	FOR(i, 1, 4)
	{
		if (dist(cur, vals[i]) < dist(cur, vals[ans]))
		{
			ans = i;
		}
	}
	return ans + 1;
	//calculate the minimum
}
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