제출 #606477

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
606477		cheissmart	팀들 (IOI15_teams)	C++14	77 / 100	3270 ms	137608 KiB

teams

#include "teams.h"
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#define F first
#define S second
#define V vector
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define SZ(v) int((v).size())
#define ALL(v) (v).begin(), (v).end()
 
using namespace std;
 
typedef long long ll;
typedef pair<int, int> pi;
typedef V<int> vi;
 
const int INF = 1e9 + 7, N = 5e5 + 7;
 
int n;
int l[N], r[N];
vi p, ev[N];

namespace pseg {
	const int MX = 3e6;
	int h[N], sz = 1;
	int tree[MX][3];
	void pull(int u) {
		tree[u][2] = tree[tree[u][0]][2] + tree[tree[u][1]][2];
	}
	int upd(int t, int pos, int tl = 0, int tr = n) {
		if(tr - tl == 1) {
			tree[sz][2] = 1;
			return sz++;
		}
		int u = sz++;
		tree[u][0] = tree[t][0], tree[u][1] = tree[t][1];
		int tm = (tl + tr) / 2;
		if(pos < tm) tree[u][0] = upd(tree[t][0], pos, tl, tm);
		else tree[u][1] = upd(tree[t][1], pos, tm, tr);
		pull(u);
		return u;
	}
	int qry(int t1, int t2, int ql, int qr, int tl = 0, int tr = n) {
		if(!t1 && !t2) return 0;
		if(ql <= tl && tr <= qr) return tree[t2][2] - tree[t1][2];
		int tm = (tl + tr) / 2;
		int ans = 0;
		if(ql < tm) ans += qry(tree[t1][0], tree[t2][0], ql, qr, tl, tm);
		if(qr > tm) ans += qry(tree[t1][1], tree[t2][1], ql, qr, tm, tr);
		return ans;
	}
	int find_who(int t1, int t2, int need) {
		int tl = 0, tr = n;
		while(tr - tl > 1) {
			int tm = (tl + tr) / 2;
			if(tree[tree[t2][0]][2] - tree[tree[t1][0]][2] > need)
				t1 = tree[t1][0], t2 = tree[t2][0], tr = tm;
			else
				need -= (tree[tree[t2][0]][2] - tree[tree[t1][0]][2]), t1 = tree[t1][1], t2 = tree[t2][1], tl = tm;
		}
		return tl;
	}
}

void init(int _n, int _l[], int _r[]) {
	n = _n;
	for(int i = 0; i < n; i++)
		l[i] = _l[i], r[i] = _r[i];
 
	p = vi(n); iota(ALL(p), 0);
	sort(ALL(p), [&] (int x, int y) {
		return r[x] < r[y];
	});
	for(int i = 0; i < n; i++)
		ev[l[p[i]]].PB(i);
	for(int i = 1; i <= n; i++) {
		pseg::h[i] = pseg::h[i - 1];
		for(int j:ev[i])
			pseg::h[i] = pseg::upd(pseg::h[i], j);
	}
}
 
struct seg {
	int r, hi;
};
 
int can(int m, int a[]) {
	if(accumulate(a, a + m, 0LL) > n) return 0;
	sort(a, a + m);
 
	V<seg> stk;
	stk.PB({0, n});
	for(int i = 0, j = 0; i < m; i++) {
		int need = 0, ii = i;
		while(i < m && a[i] == a[ii]) need += a[i++];
		i--;
 
		while(j < n && r[p[j]] < a[i]) j++;
		while(stk.back().hi < j) {
			assert(SZ(stk));
			stk.pop_back();
		}
		int at_least = j;
		while(need && SZ(stk)) {
			int lb = stk.back().r, rb = a[i]; // (lb, rb]
			int cnt = pseg::qry(pseg::h[lb], pseg::h[rb], at_least, stk.back().hi);
			if(cnt < need) {
				at_least = stk.back().hi;
				stk.pop_back();
				need -= cnt;
			} else if(cnt == need) {
				at_least = stk.back().hi;
				stk.pop_back();
				stk.PB({a[i], at_least});
				need -= cnt;
				break;
			} else {
				int dead = pseg::qry(pseg::h[lb], pseg::h[rb], 0, at_least);
				int who = pseg::find_who(pseg::h[lb], pseg::h[rb], need + dead);
				stk.PB({a[i], who});
				break;
			}
		}
		if(stk.empty()) return 0;
	}
	return 1;
}

• Subtask 1: 21 / 21
• Subtask 2: 13 / 13
• Subtask 3: 43 / 43
• Subtask 4: 0 / 23