제출 #1328072

#제출 시각아이디문제언어결과실행 시간메모리
1328072Zbyszek99Sightseeing in Kyoto (JOI22_kyoto)C++20
100 / 100
20 ms5656 KiB
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#define ll long long
#define ld long double
#define ull unsigned long long
#define ff first
#define ss second
#define pii pair<int,int>
#define pll pair<long long, long long>
#define vi vector<int>
#define vl vector<long long>
#define pb push_back
#define rep(i, b) for(int i = 0; i < (b); ++i)
#define rep2(i,a,b) for(int i = a; i <= (b); ++i)
#define rep3(i,a,b,c) for(int i = a; i <= (b); i+=c)
#define count_bits(x) __builtin_popcountll((x))
#define all(x) (x).begin(),(x).end()
#define siz(x) (int)(x).size()
#define forall(it,x) for(auto& it:(x))
using namespace __gnu_pbds;
using namespace std;
typedef tree<int, null_type, less<int>, rb_tree_tag,tree_order_statistics_node_update> ordered_set;
//mt19937 mt;void random_start(){mt.seed(chrono::time_point_cast<chrono::milliseconds>(chrono::high_resolution_clock::now()).time_since_epoch().count());}
//ll los(ll a, ll b) {return a + (mt() % (b-a+1));}
const int INF = 1e9+50;
const ll INF_L = 1e18+40;
const ll MOD = 1e9+7;

vi get_hull2(vl a)
{

    vi ans;
    rep(i,siz(a))
    {
        while(siz(ans) >= 2 && (ll)(i-ans.back())*(a[ans[siz(ans)-2]]-a[ans.back()])-(ll)(ans[siz(ans)-2]-ans.back())*(a[i]-a[ans.back()]) < 0) ans.pop_back();
        ans.pb(i);
    }
    return ans;
}

vi get_hull(vl a)
{
    pll min_ = {1e18,-1};
    rep(i,siz(a)) min_ = min(min_,{a[i],i});
    vl b,c;
    rep(i,min_.ss+1) b.pb(a[i]);
    rep2(i,min_.ss,siz(a)-1) c.pb(a[i]);
    reverse(all(c));
    vi c1 = get_hull2(b);
    vi c2 = get_hull2(c);
    c2.pop_back();
    reverse(all(c2));
    rep(i,siz(c2)) c1.pb(siz(c)-c2[i]-1+siz(b)-1);
    return c1;
}

int main()
{
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    //random_start();
    int n,m;
    cin >> n >> m;
    vl arr1(n);
    vl arr2(m);
    rep(i,n) cin >> arr1[i];
    rep(i,m) cin >> arr2[i];
    vi c1 = get_hull(arr1);
    vi c2 = get_hull(arr2);
    ll ans = 0;
    ll cur_row = 0;
    ll cur_col = 0;
    int p1 = 0;
    int p2 = 0;
    while(cur_row != n-1 || cur_col != m-1)
    {
        while(c1[p1] < cur_row) p1++;
        while(c2[p2] < cur_col) p2++;
        bool is_A = 0;
        if(cur_col == m-1) is_A = 1;
        else
        {
            if(cur_row != n-1 && (arr1[c1[p1+1]]-arr1[c1[p1]])*(c2[p2+1]-c2[p2]) <= (arr2[c2[p2+1]]-arr2[c2[p2]])*(c1[p1+1]-c1[p1])) is_A = 1;
        }
        if(is_A)
        {   
            ans += (c1[p1+1]-cur_row)*arr2[cur_col];
            cur_row = c1[p1+1];
        }
        else
        {
            ans += (c2[p2+1]-cur_col)*arr1[cur_row];
            cur_col = c2[p2+1];
        }
    }
    cout << ans << "\n";
}
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