Submission #132822

#	Time	Username	Problem	Language	Result	Execution time	Memory
132822		youngyojun	Dragon 2 (JOI17_dragon2)	C++11	60 / 100	4003 ms	7292 KiB

dragon2

#include <bits/stdc++.h>
#define eb emplace_back
#define sz(V) ((int)(V).size())
#define allv(V) ((V).begin()),((V).end())
#define sorv(V) sort(allv(V))
#define rb(x) ((x)&(-(x)))
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
pll operator + (const pll &a, const pll &b) { return pll(a.first+b.first, a.second+b.second); }
pll operator - (const pll &a, const pll &b) { return pll(a.first-b.first, a.second-b.second); }
ll operator * (const pll &a, const pll &b) { return a.first*b.second - b.first*a.second; }
ll ccw(const pll &a, const pll &b, const pll &c) { return a*b + b*c + c*a; }

const int MAXN = 30005;
const int MAXK = 30005;
const int MAXQ = 100005;
const int MX = 60055;

struct EVT {
	EVT(int p, int x, int i) : p(p), x(x), i(i) {}
	int p, x, i;

	bool operator < (const EVT &t) const {
		if(x != t.x) return x < t.x;
		return p < t.p;
	}
};

struct BIT {
	int d[MX];
	void init() { memset(d, 0, MX<<2); }
	void init(int x) {
		for(x += 5; x < MX; x += rb(x))
			d[x] = 0;
	}
	void upd(int x, int r) {
		for(x += 5; x < MX; x += rb(x))
			d[x] += r;
	}
	int get(int x) {
		int r = 0; for(x += 5; x; x -= rb(x))
			r += d[x];
		return r;
	}
	int get(int s, int e) { return s > e ? 0 : get(e)-get(s-1); }
} bit;

pll P[MAXN*2];
int A[MAXN], O[MAXN*2];

vector<int> AV[MAXK];
int L[MAXN], LS[MAXN], LE[MAXN];
int R[MAXN], RS[MAXN], RE[MAXN];
int B[MAXQ], C[MAXQ];

int Ans[MAXQ];
int N, K, Q;

void cal(int a, int b, int &ret) {
	vector<EVT> EV;
	for(int v : AV[b]) EV.eb(0, L[v], v);
	for(int v : AV[a]) {
		EV.eb(1, LS[v]-1, v);
		EV.eb(2, LE[v], v);
	}
	sorv(EV);
	for(auto &ev : EV) {
		int t = ev.p, i = ev.i;
		if(!t) bit.upd(R[i], 1);
		else if(1 == t) ret -= bit.get(RS[i], RE[i]);
		else ret += bit.get(RS[i], RE[i]);
	}
	for(int v : AV[b]) bit.init(R[v]);
}

void precal() {
	iota(O, O+(N<<1), 4);
	for(int i = 0; i <= N+1; i++)
		P[i<<1|1] = P[0] + P[0] - P[i<<1];
	sort(O, O+(N<<1), [&](int a, int b) {
		bool af = ccw(P[3], P[0], P[a]) < 0;
		bool bf = ccw(P[3], P[0], P[b]) < 0;
		if(af != bf) return af;
		return ccw(P[a], P[0], P[b]) < 0;
	});
	for(int i = 0, v; i < N; i++) {
		v = O[i];
		L[(v>>1)-1] = i + ((v&1) ? N : 0);
		v >>= 1; v--;
		LS[v] = i; LE[v] = i+N;
	}
	for(int i = 0; i <= N+1; i++)
		P[i<<1|1] = P[2] + P[2] - P[i<<1];
	sort(O, O+(N<<1), [&](int a, int b) {
		bool af = ccw(P[1], P[2], P[a]) < 0;
		bool bf = ccw(P[1], P[2], P[b]) < 0;
		if(af != bf) return af;
		return ccw(P[a], P[2], P[b]) < 0;
	});
	for(int i = 0, v; i < N; i++) {
		v = O[i];
		R[(v>>1)-1] = i + ((v&1) ? N : 0);
		v >>= 1; v--;
		RS[v] = i; RE[v] = i+N;
	}
	for(int i = 1; i <= N; i++) {
		L[i]++; LS[i]++; LE[i]++;
		R[i]++; RS[i]++; RE[i]++;
	}
}

int main() {
	ios::sync_with_stdio(false);

	cin >> N >> K;
	for(int i = 1; i <= N; i++) {
		cin >> P[(i+1)<<1].first >> P[(i+1)<<1].second >> A[i];
		AV[A[i]].eb(i);
	}
	cin >> P[0].first >> P[0].second >> P[2].first >> P[2].second;
	cin >> Q;
	for(int i = 1; i <= Q; i++) cin >> B[i] >> C[i];

	precal();

	for(int i = 1; i <= Q; i++) cal(B[i], C[i], Ans[i]);

	for(int i = 1; i <= Q; i++) printf("%d\n", Ans[i]);
	return 0;
}

• Subtask 1: 15 / 15
• Subtask 2: 45 / 45
• Subtask 3: 0 / 40