제출 #838418

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
838418		ballbattle	Aliens (IOI16_aliens)	C++14	60 / 100	2045 ms	19664 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 __float128 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 y() {return make_pair(c + (curr_x*slope), my_k);}
    //pfl operator()(ll x) {return make_pair(c + ((f)x)*slope, my_k);}
    pfl operator()(f x) {return make_pair(c + 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, ll l, ll r, ll idx);
pfl query(ll x, ll l, ll r, ll idx, f curr_x);

ll n;
ll k;

ll s;

vpll p;
vll h;

vp dp;
//f curr_x;

vl tree;
line inf;

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;}

    // Prepare Li Chao tree
    inf.c = f(10000000000000);
    inf.slope = f(0);
    s = 1;
    ll tempppp = n-1;
    for(; tempppp > 0; tempppp>>=1) {s <<= 1;}

    /*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) {
    // Initialize the Li Chao Tree
    tree.clear();
    tree.resize(2*s,inf);

    dp.clear();
    dp.resize(n,mp(f(10000000000000),0));
    //dp[0] = mp(f( (p[0].first-p[0].second+1)*(p[0].first-p[0].second+1) ),1);
    dp[0] = mp(f(0),0);
    //cout << "built xd" << endl;
    rep(i,0,n) {
        /*rep(j,0,i) {
            ll x_i = p[i].first;
            ll x_j = p[j].first;
            if (h[j] < 0) {
                dp[i] = min(dp[i], dp[j] + mp(f( (x_i-x_j+h[j]+1)*(x_i-x_j+h[j]+1) ),1) + mp(lambda,0));
            } else {
                dp[i] = min(dp[i], dp[j] + mp(f( (x_i-x_j+h[j]+1)*(x_i-x_j+h[j]+1) - (h[j]+1)*(h[j]+1) ),1) + mp(lambda,0));
            }
        }*/
        f curr_x = (f)p[i].first;
        if (i!=0) dp[i] = mp(curr_x*curr_x,0) + query(i, 0, n, 1, curr_x) + mp(lambda,1); // wtf based
        //cout << "i: " << i << " " << dp[i].first << " " << dp[i].second << endl;

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

void insert(line seg, ll l, ll r, ll idx) {
    ll m = (l+r)/2;
    f curr_x = (f)p[m].first;
    if (l+1==r) {
        if (tree[idx](curr_x) > seg(curr_x)) {swap(tree[idx],seg);}
        return;
    }
    // make seg the worse one (ie the larger one)
    if (tree[idx](curr_x) > seg(curr_x)) {swap(tree[idx],seg);}
    // no early returns bc whatever
    if (tree[idx].slope > seg.slope) {
        // The better (smaller) one has larger slope, which means the worse (larger) will undercut it later. Hence rc
        insert(seg, m, r, 2*idx + 1); // rc
    } else {
        // The better (smaller) one has smaller slopw, which means the worse (larger) will undercut it before. Hence lc
        insert(seg, l, m, 2*idx); // lc
    }
}

pfl query(ll x, ll l, ll r, ll idx, f curr_x) {
    ll m = (l+r)/2;
    if (l+1 == r) {return tree[idx](curr_x);}
    if (x < m) {
        return min(tree[idx](curr_x), query(x, l, m, 2*idx, curr_x)); // lc
    } else {
        return min(tree[idx](curr_x), query(x, m, r, 2*idx + 1, curr_x)); // rc
    }
}

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