結果

問題 No.738 平らな農地
ユーザー apricity
提出日時 2023-03-02 23:58:45
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 86 ms / 2,000 ms
コード長 3,900 bytes
コンパイル時間 1,224 ms
コンパイル使用メモリ 137,684 KB
最終ジャッジ日時 2025-02-11 01:17:26
ジャッジサーバーID
（参考情報） 		judge5 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 5
other AC * 87
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<numeric>
#include<cmath>
#include<utility>
#include<tuple>
#include<cstdint>
#include<cstdio>
#include<iomanip>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<deque>
#include<unordered_map>
#include<unordered_set>
#include<bitset>
#include<cctype>
#include<chrono>
#include<random>
#include<cassert>
#include<cstddef>
#include<iterator>
#include<string_view>
#include<type_traits>
#ifdef LOCAL
#  include "debug_print.hpp"
#  define debug(...) debug_print::multi_print(#__VA_ARGS__, __VA_ARGS__)
#else
#  define debug(...) (static_cast<void>(0))
#endif
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
#define rrep(i,n) for(int i=(n)-1; i>=0; i--)
#define FOR(i,a,b) for(int i=(a); i<(b); i++)
#define RFOR(i,a,b) for(int i=(b-1); i>=(a); i--)
#define ALL(v) v.begin(), v.end()
#define RALL(v) v.rbegin(), v.rend()
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );
#define pb push_back
using ll = long long;
using D = double;
using LD = long double;
using P = pair<int, int>;
template<typename T> using PQ = priority_queue<T,vector<T>>;
template<typename T> using minPQ = priority_queue<T, vector<T>, greater<T>>;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
void yesno(bool flag) {cout << (flag?"Yes":"No") << "\n";}
template<typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
    os << p.first << " " << p.second;
    return os;
}
template<typename T, typename U>
istream &operator>>(istream &is, pair<T, U> &p) {
    is >> p.first >> p.second;
    return is;
}
template<typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
    int s = (int)v.size();
    for (int i = 0; i < s; i++) os << (i ? " " : "") << v[i];
    return os;
}
template<typename T>
istream &operator>>(istream &is, vector<T> &v) {
    for (auto &x : v) is >> x;
    return is;
}
void in() {}
template<typename T, class... U>
void in(T &t, U &...u) {
    cin >> t;
    in(u...);
}
void out() { cout << "\n"; }
template<typename T, class... U, char sep = ' '>
void out(const T &t, const U &...u) {
    cout << t;
    if (sizeof...(u)) cout << sep;
    out(u...);
}
void outr() {}
template<typename T, class... U, char sep = ' '>
void outr(const T &t, const U &...u) {
    cout << t;
    outr(u...);
}
template<typename T>
int compress(vector<T> &v){
    vector<T> aux = v;
    sort(ALL(aux));
    UNIQUE(aux);
    rep(i,v.size()) v[i] = lower_bound(ALL(aux), v[i]) - aux.begin();
    return aux.size();
}
#include "atcoder/segtree.hpp"
ll op(ll a, ll b) {return a+b;}
ll e() {return 0;}
int half;
bool f(ll cnt){
    return cnt <= half;
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int n,k; in(n,k);
    half = k/2;
    vector<ll> a(n); in(a);
    vector<ll> a_comp = a;
    int m = compress(a_comp);
    vector<ll> idx_to_val(m);
    rep(i,n) idx_to_val[a_comp[i]] = a[i];
    atcoder::segtree<ll,op,e> cnt(m), sum(m);
    rep(i,k){
        cnt.set(a_comp[i], cnt.get(a_comp[i])+1);
        sum.set(a_comp[i], sum.get(a_comp[i])+a[i]);
    }
    auto calc_score = [&] () -> ll{
        int med_idx = cnt.max_right<f>(0);
        ll med = idx_to_val[med_idx];
        ll res = med * cnt.prod(0,med_idx) - sum.prod(0,med_idx) + sum.prod(med_idx, m) - med * cnt.prod(med_idx, m);
        return res;
    };
    ll ans = 1e18;
    rep(i,n-k+1){
        chmin(ans, calc_score());
        if(i == n-k) break;
        ll idx_out = a_comp[i];
        ll idx_in = a_comp[i+k];
        cnt.set(idx_out, cnt.get(idx_out)-1);
        cnt.set(idx_in, cnt.get(idx_in)+1);
        sum.set(idx_out, sum.get(idx_out)-a[i]);
        sum.set(idx_in, sum.get(idx_in)+a[i+k]);
    }
    out(ans);
}
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