Submission #954732

#TimeUsernameProblemLanguageResultExecution timeMemory
954732GrindMachineShortcut (IOI16_shortcut)C++17
23 / 100
2024 ms440 KiB
#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;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*



*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

#include "shortcut.h"

long long find_shortcut(int n, std::vector<int> a_, std::vector<int> b_, int c_)
{
    vector<ll> a,b;
    trav(x,a_) a.pb(x);
    trav(x,b_) b.pb(x);
    ll c = c_;

    a.insert(a.begin(),0);
    a.pb(0);
    b.insert(b.begin(),0);
    vector<ll> p(n+5);
    rep1(i,n-1){
        p[i+1] = p[i]+a[i];
    }

    vector<ll> smx(n+5);
    rep1(i,n){
        smx[i] = max(b[i],a[i-1]+smx[i-1]);
    }

    vector<ll> pmx(n+5);
    rev(i,n,1){
        pmx[i] = max(b[i],a[i]+pmx[i+1]);
    }

    auto f = [&](ll i, ll j){
        if(i > j) swap(i,j);
        return p[j]-p[i];
    };

    ll ans = inf2;

    rep1(i,n){
        for(int j = i+1; j <= n; ++j){
            ll curr = 0;
            rep1(i,n){
                amax(curr,b[i]);
            }

            rep1(x,n){
                for(int y = x+1; y <= n; ++y){
                    ll len1 = f(x,y);
                    ll len2 = f(x,i)+c+f(j,y);
                    ll val = b[x]+b[y]+min(len1,len2);
                    amax(curr,val);
                }
            }

            amin(ans,curr);
        }
    }

    return ans;
}
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