Submission #787572

#	Time	Username	Problem	Language	Result	Execution time	Memory
787572		fatemetmhr	Shortcut (IOI16_shortcut)	C++17	31 / 100	2072 ms	3220 KiB

shortcut

//  ~ Be Name Khoda ~  //

#include "shortcut.h"
#include <bits/stdc++.h>
//#pragma GCC optimize ("O3")
//#pragma GCC target("avx2")
//#pragma GCC optimize("unroll-loops,Ofast")

using namespace std;

typedef long long ll;

#define pb       push_back
#define mp       make_pair
#define all(x)   x.begin(), x.end()
#define fi       first
#define se       second

const int maxn  =  1e6   + 10;
const int maxn5 =  3e3   + 10;
const int maxnt =  1.2e6 + 10;
const int maxn3 =  1e3   + 10;
const ll mod   =  1e9   + 7;
const int lg    =  20;
const ll  inf   =  1e18;

int n;
ll c;
vector <ll> d;
ll ps[maxn5], ps2[maxn5];
vector <ll> av[maxn5][2];

ll dis(int a, int b){
    if(a > b)
        swap(a, b);
    return ps[b] - ps[a];
}

bool check(ll lim){
    for(int i = 0; i < n; i++){
        int l = i + 1, r = n;
        bool re = true;
        for(int j = 0; j < n && re; j++) for(int k = j + 1; k < n && re; k++) if(dis(j, k) + d[j] + d[k] > lim){
            ll dis1 = dis(i, j);
            ll need = lim - c - d[j] - d[k] - dis1;
            if(need < 0)
                re = false;
            int ptl = k - (upper_bound(all(av[k][0]), need) - av[k][0].begin() - 1);
            int ptr = k + (upper_bound(all(av[k][1]), need) - av[k][1].begin() - 1);
            l = max(l, ptl);
            r = min(r, ptr);
            //cout << lim << ' ' << i << ' ' << j << ' ' << k << ' ' << l << ' ' << r << endl;
            if(l > r)
                re = false;
        }
        if(re)
            return true;
    }
    return false;
}

long long find_shortcut(int N, std::vector<int> l, std::vector<int> D, int C)
{
    n = N;
    c = C;
    for(auto u : D)
        d.pb(u);
    ps[0] = 0;
    for(int i = 1; i < n; i++)
        ps[i] = ps[i - 1] + l[i - 1];
    for(int i = 0; i < n; i++){
        for(int j = i; j >= 0; j--)
            av[i][0].pb(dis(i, j));
        for(int j = i; j < n; j++)
            av[i][1].pb(dis(i, j));
    }
    ll lo = -1, hi = mod * (n + 3);
    while(hi - lo > 1){
        ll mid = (lo + hi) >> 1;
        if(check(mid))
            hi = mid;
        else
            lo = mid;
    }
    return hi;
}

• Subtask 1: 9 / 9
• Subtask 2: 14 / 14
• Subtask 3: 8 / 8
• Subtask 4: 0 / 7
• Subtask 5: 0 / 33
• Subtask 6: 0 / 22
• Subtask 7: 0 / 4
• Subtask 8: 0 / 3