Submission #58936

#	Time	Username	Problem	Language	Result	Execution time	Memory
58936		reality	Ancient Books (IOI17_books)	C++17	100 / 100	453 ms	319672 KiB

books

#include "books.h"
#include "bits/stdc++.h"
using namespace std;
#define fi first
#define se second
#define ll long long
#define dbg(v) cerr<<#v<<" = "<<v<<'\n'
#define vi vector<int>
#define vl vector <ll>
#define pii pair<int,int>
#define mp make_pair
#define db long double
#define pb push_back
#define all(s) s.begin(),s.end()
template < class T > T smin(T &a,T b) {if (a > b) a = b;return a;}
template < class T > T smax(T &a,T b) {if (a < b) a = b;return a;}
#include "books.h"
const int oo = 1e9;
int n;
int D1[20][1 << 20];
int D2[20][1 << 20];
int Log[1 << 20];
int get1(int l,int r) {
	int k = Log[r - l + 1];
	return min(D1[k][l],D1[k][r - (1 << k) + 1]);
}
int get2(int l,int r) {
	int k = Log[r - l + 1];
	return max(D2[k][l],D2[k][r - (1 << k) + 1]);
}
long long minimum_walk(std::vector<int> p, int s) {
	n = p.size();
	int a = 0,b = n - 1;
	while (a + 1 <= s && p[a] == a)
		++a;
	while (b - 1 >= s && p[b] == b)
		--b;
	int l = s,r = s;
	ll ans = 0;
	for (int i = 0;i < n;++i)
		ans += abs(i - p[i]);
	for (int i = 0;i < n;++i) {
		D1[0][i] = D2[0][i] = p[i];
	}
	for (int t = 1;t < 20;++t)
		for (int i = 0;i + (1 << t) <= n;++i) {
			D1[t][i] = min(D1[t - 1][i],D1[t - 1][i + (1 << (t - 1))]);
			D2[t][i] = max(D2[t - 1][i],D2[t - 1][i + (1 << (t - 1))]);
		}
	for (int i = 2;i <= n;++i)
		Log[i] = Log[i / 2] + 1;
	while (a < l || r < b) {
		while (get1(l,r) < l || r < get2(l,r)) {
			smin(l,get1(l,r));
			smax(r,get2(l,r));
		}
		int cl = 0,cr = 0;
		int l1 = l,r1 = l - 1;
		int l2 = r + 1,r2 = r;
		while (a < l1) {
			--l1;
			cl += 2;
			while (l1 > get1(l1,r1))
				smin(l1,get1(l1,r1));
			if (get2(l1,r1) > r)
				break;
		}
		while (r2 < b) {
			++r2;
			cr += 2;
			while (r2 < get2(l2,r2))
				smax(r2,get2(l2,r2));
			if (get1(l2,r2) < l)
				break;
		}
		if (get2(l1,r1) > r && get1(l2,r2) < l) {
			ans += min(cl,cr);
			l = l1;
			r = r2;
		} else {
			ans += cl + cr;
			break;
		}
	}
	return ans;
}

• Subtask 1: 12 / 12
• Subtask 2: 10 / 10
• Subtask 3: 28 / 28
• Subtask 4: 20 / 20
• Subtask 5: 30 / 30