Submission #160124

#	Time	Username	Problem	Language	Result	Execution time	Memory
160124		qkxwsm	Meetings (IOI18_meetings)	C++14	60 / 100	1237 ms	393720 KiB

meetings

#include "meetings.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()
#define MAXN 131313
#define LLINF 2696969696969696969

typedef long long ll;
typedef 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;

int N, Q;
ll arr[MAXN];
pii query[MAXN];
int dp[22][MAXN];
int lt[22][MAXN], rt[22][MAXN];
int sparse[22][20][MAXN];
ll mx[5013][5013], pmx[5013][5013];
vl ans;

int qmax(int w, int l, int r)
{
	if (l > r) return 0;
	int sz = 31 - __builtin_clz(r - l + 1);
	return max(sparse[w][sz][l], sparse[w][sz][r - (1 << sz) + 1]);
}
int solve(int v, int i, int j, bool flag)
{
	if (i > j) return 0;
	if (v == 0) return 0;
	if (qmax(21, i, j) < v)
	{
		return (j - i + 1) + solve(v - 1, i, j, flag);
	}
	int res = 0;
	if (flag)
	{
		int pos = lt[v][j];
		res = qmax(v - 1, i, pos - 1);
		ckmax(res, solve(v - 1, pos + 1, j, flag));
		//answer on the right side.
	}
	else
	{
		int pos = rt[v][i];
		res = qmax(v - 1, pos + 1, j);
		ckmax(res, solve(v - 1, i, pos - 1, flag));
	}
	//solve (i...j) with value v.
	//you actually wanna figure out how much dough you save
	res += (j - i + 1);
	return res;
}

vl minimum_costs(vi h, vi l, vi r)
{
	N = SZ(h); Q = SZ(r); ans.resize(Q);
	FOR(i, 0, N)
	{
		arr[i] = h[i];
		sparse[21][0][i] = arr[i];
	}
	FOR(j, 1, 20)
	{
		FOR(k, 0, N - (1 << j) + 1)
		{
			sparse[21][j][k] = max(sparse[21][j - 1][k], sparse[21][j - 1][k + (1 << (j - 1))]);
		}
	}
	FOR(i, 0, Q)
	{
		query[i] = {l[i], r[i]};
	}
	if (N <= 5013 && Q <= 5013)
	{
		FOR(i, 0, N)
		{
			mx[i][i] = arr[i];
			FOR(j, i + 1, N)
			{
				mx[i][j] = max(mx[i][j - 1], arr[j]);
				mx[j][i] = mx[i][j];
			}
		}
		FOR(i, 0, N)
		{
			FOR(j, 0, N)
			{
				pmx[i][j + 1] = pmx[i][j] + mx[i][j];
			}
		}
		FOR(i, 0, Q)
		{
			int l = query[i].fi, r = query[i].se;
			ans[i] = LLINF;
			FOR(j, l, r + 1)
			{
				ckmin(ans[i], pmx[j][r + 1] - pmx[j][l]);
			}
		}
		return ans;
	}
	FOR(i, 1, 21)
	{
		vi guys;
		guys.PB(-1);
		FOR(j, 0, N)
		{
			if (arr[j] > i) guys.PB(j);
		}
		guys.PB(N);
		FOR(j, 1, SZ(guys))
		{
			int best = 0;
			FOR(k, guys[j - 1] + 1, guys[j])
			{
				ckmax(best, dp[i - 1][k]);
			}
			best += (guys[j] - guys[j - 1] - 1);
			FOR(k, guys[j - 1] + 1, guys[j])
			{
				ckmax(dp[i][k], best);
			}
		}
		//build the sparse table!
		FOR(j, 0, N)
		{
			sparse[i][0][j] = dp[i][j];
		}
		FOR(j, 1, 20)
		{
			FOR(k, 0, N - (1 << j) + 1)
			{
				sparse[i][j][k] = max(sparse[i][j - 1][k], sparse[i][j - 1][k + (1 << (j - 1))]);
			}
		}
	}
	FOR(i, 0, N)
	{
		FOR(j, 0, 22)
		{
			lt[j][i] = (i == 0 ? -1 : lt[j][i - 1]);
			if (arr[i] >= j) lt[j][i] = i;
		}
	}
	FORD(i, N, 0)
	{
		FOR(j, 0, 22)
		{
			rt[j][i] = (i == N - 1 ? N : rt[j][i + 1]);
			if (arr[i] >= j) rt[j][i] = i;
		}
	}
	//first occurence of >= x before y.
	FOR(i, 0, Q)
	{
		int l = query[i].fi, r = query[i].se;
		int mx = qmax(21, l, r), lf = rt[mx][l], rg = lt[mx][r];
		ans[i] = qmax(mx - 1, lf + 1, rg - 1);
		ckmax(ans[i], solve(mx - 1, l, lf - 1, 0));
		ckmax(ans[i], solve(mx - 1, rg + 1, r, 1));
		//idk solve!
		ans[i] = mx * (r - l + 1) - ans[i];
	}
	//you just store how much you can save;
	return ans;
}

• Subtask 1: 4 / 4
• Subtask 2: 15 / 15
• Subtask 3: 17 / 17
• Subtask 4: 24 / 24
• Subtask 5: 0 / 40