답안 #128452

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
128452	2019-07-10T23:59:46 Z	qkxwsm	새 집 (APIO18_new_home)	C++14	10 / 100	5000 ms	412576 KB

new_home

#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 LIM 100000007
#define INF 1000000007
#define LLINF 2696969696969696969ll
#define MAXN 300013

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<pii, pii> ppp;
typedef pair<pii, int> ppi;

int N, K, Q, S;
ppp arr[MAXN];
ppi quer[MAXN];
int ans[MAXN];
multiset<int> pts[MAXN];
vector<array<int, 4> > segments;
vector<array<int, 4> > events;
unordered_map<ll, vi> cancer;
vpi seg[MAXN << 2];
int iter[MAXN << 2];
vi compress;

void ins(int l, int r, int t)
{
	// cerr << "ins " << l << ' ' << r << " time " << t << endl;
	l = (l + r + 1) >> 1;
	cancer[1ll * INF * l + r].PB(SZ(segments));
	segments.PB({l, r, t, SZ(events) + 1});
}
void del(int l, int r, int t)
{
	// cerr << "del " << l << ' ' << r << " time " << t << endl;
	l = (l + r + 1) >> 1;
	// cerr << "rem " << (cancer[{l, r}]).size() << endl;
	segments[cancer[1ll * INF * l + r].back()][3] = t;
	cancer[1ll * INF * l + r].pop_back();
}

void inseg(int w, int L, int R, int a, int b, pii p)
{
	if (a <= L && R <= b)
	{
		seg[w].PB(p);
		return;
	}
	int mid = (L + R) >> 1;
	if (a <= mid) inseg(w << 1, L, mid, a, b, p);
	if (mid < b) inseg(w << 1 | 1, mid + 1, R, a, b, p);
	return;
}
void build(int w, int L, int R)
{
	if (L != R)
	{
		int mid = (L + R) >> 1;
		build(w << 1, L, mid);
		build(w << 1 | 1, mid + 1, R);
	}
	FOR(i, 1, SZ(seg[w]))
	{
		ckmax(seg[w][i].se, seg[w][i - 1].se);
	}
}
int query(int w, int L, int R, int a, int x)
{
	int res = -INF;
	while(iter[w] + 1 < SZ(seg[w]) && seg[w][iter[w] + 1].fi <= x)
	{
		iter[w]++;
	}
	if (iter[w] < SZ(seg[w]) && seg[w][iter[w]].fi <= x)
	{
		res = -x + seg[w][iter[w]].se;
	}
	if (L == R) return res;
	int mid = (L + R) >> 1;
	if (a <= mid) ckmax(res, query(w << 1, L, mid, a, x));
	else ckmax(res, query(w << 1 | 1, mid + 1, R, a, x));
	return res;
}

void solve()
{
	//consider distnace functions with negative slope ONLY.
	//arr: something of type t and position p in times (t1, t2)
	//quer: query of position p at time t, qid i
	FOR(i, 0, (1 << (33 - __builtin_clz(Q + 2))) + 1)
	{
		seg[i].clear();
		iter[i] = 0;
	}
	events.clear();
	cancer.clear();
	segments.clear();
	FOR(i, 0, K) pts[i].clear();
	FOR(i, 0, N)
	{
		int t1 = arr[i].se.fi, t2 = arr[i].se.se, ty = arr[i].fi.fi, x = arr[i].fi.se;
		if (t1 == t2) continue;
		events.PB({t1, -1, ty, x});
		events.PB({t2, -2, ty, x});
	}
	sort(ALL(events));
	FOR(i, 0, K)
	{
		pts[i].insert(2 * LIM);
		pts[i].insert(-2 * LIM);
		ins(-2 * LIM, 2 * LIM, 0);
	}
	FOR(i, 0, SZ(events))
	{
		int t = events[i][0], typ = events[i][2], pos = events[i][3];
		if (events[i][1] == -1)
		{
			auto it = pts[typ].UB(pos);
			int rt = *it, lt = *prev(it);
			pts[typ].insert(pos);
			ins(lt, pos, t + 1);
			ins(pos, rt, t + 1);
			del(lt, rt, t);
		}
		if (events[i][1] == -2)
		{
			pts[typ].erase(pts[typ].find(pos));
			auto it = pts[typ].UB(pos);
			int rt = *it, lt = *prev(it);
			del(lt, pos, t);
			del(pos, rt, t);
			ins(lt, rt, t + 1);
		}
	}
	sort(ALL(segments));
	FOR(i, 0, SZ(segments))
	{
		int l = segments[i][2], r = segments[i][3]; if (l > r) continue;
		inseg(1, 0, Q + 2, l, r, {segments[i][0], segments[i][1]});
	}
	//ok cool we have the segments
	build(1, 0, Q + 2);
	FOR(i, 0, Q)
	{
		ckmax(ans[quer[i].se], query(1, 0, Q + 2, quer[i].fi.se + 1, quer[i].fi.fi));
	}
}
bool cmp(ppi a, ppi b)
{
	return a.fi.se < b.fi.se;
}

int32_t main()
{
	if (fopen("file.in", "r"))
	{
		freopen("file.in", "r", stdin);
		freopen("file.out", "w", stdout);
	}
	ios_base::sync_with_stdio(0); cin.tie(0);
	cin >> N >> K >> Q;
	FOR(i, 0, N)
	{
		cin >> arr[i].fi.se >> arr[i].fi.fi >> arr[i].se.fi >> arr[i].se.se;
		arr[i].fi.fi--; arr[i].se.se++;
	}
	FOR(i, 0, Q)
	{
		cin >> quer[i].fi.fi >> quer[i].fi.se; quer[i].se = i;
		compress.PB(quer[i].fi.se);
	}
	compress.PB(INF);
	compress.PB(0);
	sort(ALL(compress)); compress.erase(unique(ALL(compress)), compress.end());
	FOR(i, 0, Q)
	{
		quer[i].fi.se = LB(ALL(compress), quer[i].fi.se) - compress.begin();
	}
	FOR(i, 0, N)
	{
		arr[i].se.fi = LB(ALL(compress), arr[i].se.fi) - compress.begin();
		arr[i].se.se = LB(ALL(compress), arr[i].se.se) - compress.begin();
	}
	sort(arr, arr + N);
	sort(quer, quer + Q);
	solve();
	FOR(i, 0, N)
	{
		arr[i].fi.se = 100000001 - arr[i].fi.se;
	}
	FOR(i, 0, Q)
	{
		quer[i].fi.fi = 100000001 - quer[i].fi.fi;
	}
	reverse(quer, quer + Q);
	solve();
	//events: OPEN/CLOSE a store at time t, position p, type x, ASK at time t, position p
	FOR(i, 0, Q)
	{
		if (ans[i] >= LIM)
		{
			cout << "-1\n";
			continue;
		}
		// cerr << "ans " << ans[i] << endl;
		cout << ans[i] << '\n';
	}
	// cerr << "time elapsed = " << clock() / ((CLOCKS_PER_SEC) / 1000) << " ms" << endl;
	return 0;
}

Compilation message

new_home.cpp: In function 'int32_t main()':
new_home.cpp:185:10: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
   freopen("file.in", "r", stdin);
   ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~
new_home.cpp:186:10: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
   freopen("file.out", "w", stdout);
   ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~

Subtask #1
0 / 5.0

#	결과	실행 시간	메모리	Grader output
1	Correct	41 ms	42616 KB	Output is correct
2	Correct	40 ms	42616 KB	Output is correct
3	Incorrect	41 ms	42616 KB	Output isn't correct
4	Halted	0 ms	0 KB	-

Subtask #2
0 / 7.0

#	결과	실행 시간	메모리	Grader output
1	Correct	41 ms	42616 KB	Output is correct
2	Correct	40 ms	42616 KB	Output is correct
3	Incorrect	41 ms	42616 KB	Output isn't correct
4	Halted	0 ms	0 KB	-

Subtask #3
10.0 / 10.0

#	결과	실행 시간	메모리	Grader output
1	Correct	3868 ms	285732 KB	Output is correct
2	Correct	3891 ms	261416 KB	Output is correct
3	Correct	3961 ms	407176 KB	Output is correct
4	Correct	3755 ms	274944 KB	Output is correct
5	Correct	3942 ms	256448 KB	Output is correct
6	Correct	4139 ms	261568 KB	Output is correct
7	Correct	4259 ms	412576 KB	Output is correct
8	Correct	3761 ms	289048 KB	Output is correct
9	Correct	3765 ms	269952 KB	Output is correct
10	Correct	4030 ms	262652 KB	Output is correct
11	Correct	3757 ms	257036 KB	Output is correct
12	Correct	3543 ms	262572 KB	Output is correct