제출 #144484

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
144484		liwi	Aliens (IOI16_aliens)	C++11	4 / 100	3 ms	1148 KiB

aliens

#include "aliens.h"

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag,tree_order_statistics_node_update> ordered_set;

#define scan(x) do{while((x=getchar())<'0'); for(x-='0'; '0'<=(_=getchar()); x=(x<<3)+(x<<1)+_-'0');}while(0)
char _;
#define complete_unique(a) a.erase(unique(a.begin(),a.end()),a.end())
#define all(a) a.begin(),a.end()
#define println printf("\n");
#define readln(x) getline(cin,x);
#define pb push_back
#define endl "\n"
#define INT_INF 0x3f3f3f3f
#define LL_INF 0x3f3f3f3f3f3f3f3f
#define MOD 1000000007
#define MOD2 1190492669
#define SEED 131
#define mp make_pair
#define fastio cin.tie(0); cin.sync_with_stdio(0);

#define MAXN 50005
#define MAXK 105

typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
typedef unordered_map<int,int> umii;
typedef pair<int,int> pii;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef pair<int,pii> triple;
typedef int8_t byte;

mt19937 g1(time(0));

int randint(int a, int b){return uniform_int_distribution<int>(a, b)(g1);}
ll randlong(ll a,ll b){return uniform_int_distribution<long long>(a, b)(g1);}

ll gcd(ll a, ll b){return b == 0 ? a : gcd(b, a % b);}
ll lcm(ll a, ll b){return a*b/gcd(a,b);}
ll fpow(ll  b, ll exp, ll mod){if(exp == 0) return 1;ll t = fpow(b,exp/2,mod);if(exp&1) return t*t%mod*b%mod;return t*t%mod;}
ll divmod(ll i, ll j, ll mod){i%=mod,j%=mod;return i*fpow(j,mod-2,mod)%mod;}

int num_points,len,num_photo,ptr,cnt[MAXN];
pii points[MAXN],p[MAXN],lst;
ld dp[MAXN];
ll m[MAXN];

inline bool cmp(pii &a, pii &b){
	if(a.first == b.first) return a.second>b.second;
	return a.first<b.first;
}

inline ld calc(int i, int j){
	ld c1 = dp[j]+(ld)(p[i].second-p[j+1].first+1)*(p[i].second-p[j+1].first+1);
	ld c2 = m[j]*m[j];
	return c1-c2;
}

inline ld inter(int j, int k){
	ld m1 = (ld)-2.0*(p[j+1].first-1), b1 = (ld)(p[j+1].first-1)*(p[j+1].first-1)+dp[j]-m[j]*m[j];
	ld m2 = (ld)-2.0*(p[k+1].first-1), b2 = (ld)(p[k+1].first-1)*(p[k+1].first-1)+dp[k]-m[k]*m[k];
	return (ld)(b2-b1)/(m1-m2);
}

inline int getGroups(ld c){
	deque<int> dq; dq.pb(0);
//		for(int i=0; i<=ptr; i++){
//			if(dp[k-1][i] == LL_INF) continue;
//			while(dq.size() >= 2 && inter(dq[dq.size()-2],i,k) <= inter(dq[dq.size()-2],dq.back(),k)) dq.pop_back();
//			dq.pb(i);
//		}
	//dp[k][i] = max(dp[k-1][v]+(p[i]-(p[v+1]-1))*(p[i]-(p[v+1]-1))-m[v]*m[v]
	//dp[k][i] = dp[k-1][v]+p[i]^2-2*p[i]*(p[v+1]-1)+(p[v+1]-1)^2-m[v]^2
	//dp[k][i] = -2*(p[v+1]-1)*p[i]+(p[v+1]-1)^2+dp[k-1][v]-m[v]^2
	//m = -2*(p[v+1]-1);
	//b = (p[v+1]-1)^2+dp[k-1][v]-m[v]^2
	dp[0] = 0;
	for(int i=1; i<=ptr; i++){
		while(dq.size() >= 2 && calc(i,dq[0]) >= calc(i,dq[1])) dq.pop_front();
		dp[i] = calc(i,dq[0])-c;
		cnt[i] = cnt[dq[0]]+1;
		while(dq.size() >= 2 && inter(dq[dq.size()-2],i) <= inter(dq[dq.size()-2],dq.back())) dq.pop_back();
		dq.pb(i);
//			for(int v=i-1; v>=0; v--){
//				ll cost = dp[k-1][v]+1LL*(p[i].second-p[v+1].first+1)*(p[i].second-p[v+1].first+1);
//				ll overlap = m[v]*m[v];
//				dp[k][i] = min(dp[k][i],cost-overlap);
//			}
	}
	return cnt[ptr];
}

ll take_photos(int nn, int mm, int kk, vector<int> rows, vector<int> cols){
	num_points = nn, len = mm, num_photo = kk, ptr = 0;
	for(int i=1; i<=num_points; i++){
		int a = min(rows[i-1],cols[i-1]);
		int b = max(rows[i-1],cols[i-1]);
		points[i] = mp(a,b);
	}
	sort(points+1,points+num_points+1,cmp);
	lst = p[++ptr] = points[1], p[0] = mp(-1,-1);
	for(int i=2; i<=num_points; i++){
		int r = points[i].second;
		if(r > lst.second){
			p[++ptr] = points[i];
			lst = points[i];
		}
	}
	for(int i=1; i<ptr; i++)
		m[i] = max(0,p[i].second-p[i+1].first+1);
	memset(dp,0x3f,sizeof dp);
	ld low = 0, high = 1e18, ans = 0;
	while(high-low >= 1e-18){
		ld mid = (low+high)/2;
		if(getGroups(mid) <= num_photo){
			ans = mid;
			high = mid-1e-18;
		}else low = mid+1e-18;
	}
	getGroups(ans);
	ll res = (ll)(dp[ptr]+ans*num_photo);
	return res;
}

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