제출 #138849

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
138849		mosesmayer	Aliens (IOI16_aliens)	C++17	4 / 100	4 ms	2040 KiB

aliens

#include "aliens.h"
#include <bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;

struct Line{
	LL m, c;
	Line(LL m = 0, LL c = 0): m(m), c(c){}
	LL get(LL x){return m * x + c;}
};
struct ConvexHull{
	int sz, B;
	Line *hull;
	int *idx;
	ConvexHull(int n): sz(0), B(0){
		hull = new Line[++n];
		idx = new int[n];
	}
	bool is_bad(int curr, int prev, int next){
		Line c = hull[curr], p = hull[prev], n = hull[next];
		return (c.c - n.c) * (c.m - p.m) <= (p.c - c.c) * (n.m - c.m);
	}
	void add_line(LL m, LL c, int id){
		hull[sz++] = Line(m, c);
		idx[sz] = id;
		while (sz > 2 && is_bad(sz-2, sz-3, sz-1)){
			hull[sz-2] = hull[sz-1]; idx[sz-2] = idx[sz-1]; sz--;
		}
		B = min(B, sz - 1);
	}

	pair<LL, int> query(LL x){
		while (B < sz - 1 && hull[B].get(x) >= hull[B+1].get(x)) B++;
		return make_pair(hull[B].get(x), idx[B]);
	}
};

const LL LINF = 1e18;
const int MX = 1e5;
int n, m, k;
vector<pii> ori;
LL L[MX + 5], R[MX + 5], nx[MX + 5];
LL dp[MX + 5];
int cnt[MX + 5];
ConvexHull ch(MX + 5);

bool cmp(pii a, pii b){
	return a.fi != b.fi ? a.fi < b.fi : a.se > b.se;
}

/*
   dp[k][n] = min(dp[k-1][n], min{j<=n} (dp[k-1][j-1] + (L[j]^2-2L[j]) + max(0,R[j-1]-L[j]+1)^2 - 2L[j]•R[n]) + R[n]^2 + 2R[n] + 1;
										(             c1             )   (      c2            ) (  m  ) ( x )   (   extra        )
*/

int check(LL C){
	ch.sz = 0; ch.B = 0;
	for (int j = 1; j <= n; j++){
		ch.add_line(-2LL * L[j], dp[j-1] + L[j] * (L[j] - 2) - nx[j] * nx[j], j);
		pair<LL, int> res = ch.query(R[j]);
		dp[j] = res.first + R[j] * R[j] + 2LL * R[j] + 1 + C;
		cnt[j] = cnt[res.second] + 1;
//		cout << "J/" << j << " : " << dp[j] << ' ' << cnt[j] << '\n';
	}
//	cout << "Checking C : " << C << " with result " << dp[n] << ' ' << cnt[n] << '\n';
	return cnt[n];
}
long long take_photos(int N, int M, int K, std::vector<int> r, std::vector<int> c) {
	n = N, m = M, k = K;
	for (int i = 0; i < n; i++){
		ori.emplace_back(r[i], c[i]);
		if (ori.back().fi > ori.back().se) swap(ori.back().fi, ori.back().se);
	}
	sort(ori.begin(), ori.end());
	int idx = 0;
	for (int i = 0; i < n; i++){
		if (idx && L[idx] <= ori[i].fi && R[idx] >= ori[i].se) continue;
		while (idx && ori[i].fi <= L[idx] && ori[i].se >= R[idx]) idx--;
		idx++; tie(L[idx], R[idx]) = ori[i];
	}
	n = idx;
//	for (int i = 1; i <= n; i++) printf("(%d, %d)%c", L[i], R[i], " \n"[i==n]);

	L[0] = R[0] = -1;
	for (int i = 1; i <= n; i++)
		nx[i] = max(0LL, R[i-1] - L[i] + 1);
//	for (int i = 1; i <= n; i++) printf("%lld%c", nx[i], " \n"[i==n]);
	LL lo = 0, hi = 1e18, md, ans = 0;
	while (lo <= hi){
		md = (lo + hi) >> 1;
		if (check(md) >= k){
			ans = md; lo = md + 1;
		} else hi = md - 1;
	}
	int kk = check(ans);
//	cout << ans << ' ' << kk << '\n';
	return dp[n] - ans * kk;
}

• Subtask 1: 4 / 4
• Subtask 2: 0 / 12
• Subtask 3: 0 / 9
• Subtask 4: 0 / 16
• Subtask 5: 0 / 19
• Subtask 6: 0 / 40