제출 #338154

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
338154		cki86201	고대 책들 (IOI17_books)	C++17	50 / 100	381 ms	32340 KiB

books

#include <bits/stdc++.h>
#include "books.h"
using namespace std;

#define Fi first
#define Se second
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define szz(x) (int)(x).size()
#define rep(i, n) for(int i=0;i<(n);i++)
typedef tuple<int, int, int> t3;

int c[1000010], lv[1000010], rv[1000010], T[1000010], S[1000010];
ll minimum_walk(vector<int> p, int s) {
	int n = szz(p);
	ll ans = 0;
	rep(i, n) ans += abs(i - p[i]);
	auto upd = [&](int s, int e) {
		for(int i=s;i<=e;i++) {
			T[i] += (i > s ? T[i-1] : 0);
			S[i] = (i > s ? S[i-1] : 0) + (T[i] == 0);
		}
	};
	auto clr = [&](int s, int e) {
		for(int i=s;i<=e;i++) T[i] = S[i] = 0;
	};
	auto gs = [&](int s, int e) {
		return (e ? S[e-1] : 0) - (s ? S[s-1] : 0);
	};
	rep(i, n) if(!c[i]) {
		int l = i, r = i;
		for(int j=p[i];j!=i;j=p[j]) l = min(l, j), r = max(r, j);
		c[i] = 1, lv[i] = l, rv[i] = r;
		for(int j=p[i];j!=i;j=p[j]) c[j] = 1, lv[j] = l, rv[j] = r;
	}
	memset(c, 0, sizeof c);
	int l = s, r = s;
	auto expend = [&]() {
		for(int x=r+1;x<n;x++) {
			if(lv[x] < l) {
				int r2 = x, l2 = -1;
				for(int i=p[r2];i!=r2;i=p[i]) {
					if(i < l) l2 = max(l2, i);
				}
				return pii(l2, r2);
			}
		}
		return pii(-1, -1);
	};
	vector <int> q; q.pb(s);
	auto push = [&](int x) {
		if(l > x) {
			for(int a=l-1;a>=x;a--) q.pb(a);
			l = x;
		}
		if(r < x) {
			for(int a=r+1;a<=x;a++) q.pb(a);
			r = x;
		}
	};
	while(1) {
		rep(i, szz(q)) {
			int t = q[i];
			if(c[t]) continue;
			c[t] = 1;
			for(int x=p[t];x!=t;x=p[x]) {
				push(x);
				c[x] = 1;
			}
		}
		auto [l2, r2] = expend();
		if(l2 == -1) break;
		q.clear();
		for(int i=l2+1;i<r2;i++) T[min(i, p[i])]++, T[max(i, p[i])]--;
		upd(l2, r2);
		ans += 2 * min(gs(l2, l), gs(r, r2));
		clr(l2, r2);
		push(l2);
		push(r2);
	}
	int L = 0, R = n - 1;
	while(L < l && p[L] == L) ++L;
	while(r < R && p[R] == R) --R;
	rep(i, n) T[min(i, p[i])]++, T[max(i, p[i])]--;
	upd(0, n - 1);
	ans += 2 * (gs(L, l) + gs(r, R));
	return ans;
}

• Subtask 1: 12 / 12
• Subtask 2: 10 / 10
• Subtask 3: 28 / 28
• Subtask 4: 0 / 20
• Subtask 5: 0 / 30