제출 #838959

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
838959		ballbattle	Aliens (IOI16_aliens)	C++14	100 / 100	1488 ms	340872 KiB

aliens

#include "aliens.h"

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>

using namespace std;

typedef long long ll;
typedef long double f;

typedef pair<ll,ll> pll;
typedef pair<f,ll> pfl;

typedef vector<ll> vll;
typedef vector<pll> vpll;
typedef vector<pfl> vp;
typedef vector<int> vi;

struct line {
    f c;
    f slope;
    pfl operator()(ll x) {return make_pair(c + ((f)x)*slope, my_k);}
    ll my_k;
};
typedef vector<line* > vl;

#define rep(a,b,c) for(ll a = b; a < c; a++)
#define pb(a) push_back(a)
#define mp(a,b) make_pair(a,b)
#define all(a) a.begin(),a.end()
#define sz(a) a.size()

pfl operator+ (pfl a, pfl b) {return mp(a.first+b.first,a.second+b.second);}

pfl _dp(f lambda);
void insert(line*seg);
pfl query(f x);

ll n;
ll k;

vpll p;
vll h;

vp dp;

vl lines;

long long take_photos(int N, int M, int K, vi R, vi C) {
    k = K;
    p.clear();
    rep(i,0,N) {if (R[i] > C[i]) {swap(R[i],C[i]);}} // Flip over diagonal
    rep(i,0,N) {p.pb(mp(C[i],-R[i]));}
    sort(all(p)); // Sort in a way that makes worse points come before better points
    rep(i,0,N) {
        C[i] = p[i].first;
        R[i] = -p[i].second;
    }
    /*cout << "C[i] R[i]" << endl;
    rep(i,0,N) {cout << C[i] << " " << R[i] << endl;}*/
    // C[i] is in increasing order
    // R[i] is in decreasing order
    // The first point is the optimal

    p.clear();
    p.pb(mp(-1,-1)); // thing....
    rep(i,0,N) {

        while ((C[i] >= p[sz(p)-1].first) && (R[i] <= p[sz(p)-1].second)) { // Prune bad points
            p.pop_back();
        }
        p.pb(mp(C[i],R[i])); // Add points in (x,-y) format to p
    }
    
    /*cout << "points:" << endl;
    rep(i,0,sz(p)) {cout << p[i].first << " " << p[i].second << endl;}*/

    n = sz(p);
    h.resize(n,0);
    rep(i,0,n-1) {h[i] = p[i].first - p[i+1].second;}

    /*cout << "h:" << endl;
    rep(i,0,n) {cout << h[i] << " ";}
    cout << endl;*/

    /*rep(i,0,10) {cout << _dp(f(i)).first << endl;}*/

    ll ans = 1000000000001;

    f lo = f(0.0);
    f hi = f(1000000000001);
    f mid;
    pfl curr;
    rep(i,0,64) {
        mid = (lo)/f(2) + (hi)/f(2);
        //cout << mid << endl;
        curr = _dp(mid);
        //cout << (long double)(curr.first) << " " << curr.second << " " << (long double)(mid) << endl; 
        if (curr.second == min(k,n-1)) {
            ans = llround(curr.first - mid*curr.second);
            //cout << llround((long double)(curr.first - mid*curr.second)) << " " << (curr.first - mid*curr.second) << endl;
            break;
        } else if (curr.second > min(k,n-1)) {
            lo = mid;
        } else {
            hi = mid;
            /*ll tmp = round(curr.first - mid*(f(curr.second))); // Hotfix
            ans = min(tmp, ans);*/
        }
    }
    if (ans == 1000000000001) {ans = round(curr.first - mid*(f(k)));} // Based solution
    //if (ans==-1) {ans = round(curr.first - mid*(f(curr.second)));}

    //cout << n-1 << endl;
    return ans;
}

pfl _dp(f lambda) {
    lines.clear();
    dp.clear();
    dp.resize(n,mp((f)0,0));
    rep(i,0,n) {
        if (i!=0) dp[i] = mp((f)(p[i].first*p[i].first),0) + query((f)p[i].first) + mp(lambda,1); // wtf based

        line* new_line = new line;
        //lines.pb(new line);
        new_line->my_k = dp[i].second; // include k xd
        new_line->c = dp[i].first + (f)(p[i].first*p[i].first - 2*p[i].first*(h[i]+1) + ((h[i] < 0) ? (h[i]+1)*(h[i]+1) : 0));
        new_line->slope = (f)(-2*(p[i].first-h[i]-1));
        insert(new_line);
    }
    /*cout << "dp: ";
    rep(i,0,n) {cout << dp[i].first - lambda*dp[i].second << "," << dp[i].second << " ";}
    cout << endl;
    cout << lines.size() << ": ";
    rep(i,0,lines.size()) {cout << lines[i]->c << "," << lines[i]->slope << "," << lines[i]->my_k << " ";}
    cout << endl;*/
    return dp[n-1];
}

pfl query(f x) { // binary search
    ll lo = 0;
    ll hi = lines.size()-1;

    ll mid = (lo+hi)/2;
    while (lo != hi) {
        f intersect = (lines[mid]->c - lines[mid+1]->c)/(lines[mid+1]->slope - lines[mid]->slope);
        if (intersect < x) { // move right
            lo = mid+1;
        } else { // move left
            hi = mid;
        }
        mid = (lo+hi)/2;
    }
    return (*lines[mid])(x);
}

void insert(line*seg) { // done
    while (lines.size() >= 2) { // remove now defunct lines
        line* curr = lines.back();
        f i1 = (curr->c - seg->c)/(seg->slope - curr->slope);
        line* curr2 = lines[lines.size()-2];
        f i2 = (curr2->c - curr->c)/(curr->slope - curr2->slope);
        if (i1 < i2) {lines.pop_back();}
        else {break;}
    }
    lines.pb(seg);
}
// Insertsection of two lines
// y = x*slope + c
// x*s1 + c1 = x*s2 + c2
// x*(s1-s2) = c2-c1
// x = (c2-c1)/(s1-s2);

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