Submission #948073

#TimeUsernameProblemLanguageResultExecution timeMemory
948073steveonalexBoxes with souvenirs (IOI15_boxes)C++11
50 / 100
882 ms19136 KiB
#include <bits/stdc++.h>
#include "boxes.h"
 
using namespace std;
 
typedef long long ll;
typedef unsigned long long ull;
 
#define MASK(i) (1ULL << (i))
#define GETBIT(mask, i) (((mask) >> (i)) & 1)
#define ALL(v) (v).begin(), (v).end()
 
ll max(ll a, ll b){return (a > b) ? a : b;}
ll min(ll a, ll b){return (a < b) ? a : b;}
 
ll LASTBIT(ll mask){return (mask) & (-mask);}
int pop_cnt(ll mask){return __builtin_popcountll(mask);}
int ctz(ll mask){return __builtin_ctzll(mask);}
int logOf(ll mask){return 63 - __builtin_clzll(mask);}
 
mt19937_64 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
ll rngesus(ll l, ll r){return l + (ull) rng() % (r - l + 1);}
 
template <class T1, class T2>
    bool maximize(T1 &a, T2 b){
        if (a < b) {a = b; return true;}
        return false;
    }
 
template <class T1, class T2>
    bool minimize(T1 &a, T2 b){
        if (a > b) {a = b; return true;}
        return false;
    }
 
template <class T>
    void printArr(T& container, string separator = " ", string finish = "\n", ostream &out = cout){
        for(auto item: container) out << item << separator;
        out << finish;
    }
 
template <class T>
    void remove_dup(vector<T> &a){
        sort(ALL(a));
        a.resize(unique(ALL(a)) - a.begin());
    }

const ll INF = 1e18;
 
ll delivery(int n, int k, int L, int p[]) {
    vector<int> a;
    for(int i = 0; i<n; ++i) a.push_back(p[i]);
 
    ll ans = INF;
    for(int i = 0; i<n; i += k){
        int l = i, r = min(i + k - 1, n - 1);
        ll cur = min({a[r] * 2, L, (L - a[l]) * 2});
        ans += cur;
    }
    vector<int> valid;
    if (n % k == 0) valid = {0, k};
    else valid = {0, n % k};
    for(int t: valid){
        for(int l = -1; l < t; ++l)
        for(int r = n; (n - r) + l + 1 <= t; --r){
            ll cur = 0;
            if (l >= 0) cur += a[l] * 2;
            if (r < n) cur += (L - a[r]) * 2;
            for(int i = l + 1; i < r; i += k){
                int x = i, y = min(i + k - 1, n - 1);
                cur += min({a[y] * 2, L, (L - a[x]) * 2});
            }
            minimize(ans, cur);
        }
    }
    return ans;
}
 
 
// int main(void){
//     ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
 
 
//     return 0;
// }
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