Submission #425889

#	Time	Username	Problem	Language	Result	Execution time	Memory
425889		jtnydv25	Shortcut (IOI16_shortcut)	C++17	0 / 100	270 ms	16728 KiB

shortcut

#include "shortcut.h"
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pii pair<int, int>
#define all(c) ((c).begin()), ((c).end())
#define sz(x) ((int)(x).size())
 
#ifdef LOCAL
#include <print.h>
#else
#define trace(...)
#define endl '\n'
#endif
const int N = 1 << 20;
const ll INF = 1e18;
struct Data{
    ll a, b;
    Data() : a(-INF), b(INF){}
    Data(ll a, ll b) : a(a), b(b){}
};
 
inline void merge(Data & X, const Data & Y){
    X.a = max(X.a, Y.a);
    X.b = min(X.b,Y.b);
}
 
Data prefix[N];
ll p[N];

int perm1[N], perm2[N];
long long find_shortcut(int n, std::vector<int> l, std::vector<int> d, int c){
    for(int i = 1; i < n; i++) p[i] = p[i - 1] + l[i - 1];
    iota(perm1, perm1 + n, 0);
    iota(perm2, perm2 + n, 0);
    sort(perm1, perm1 + n, [&](int i, int j){return d[i] - p[i] > d[j] - p[j];});
    sort(perm2, perm2 + n, [&](int i, int j){return d[i] + p[i] < d[j] + p[j];});
 
    int mx = 0, mx2 = 0, ind_mx = 0;
    for(int i = 1; i <= n; i++){
        int ind = perm1[i - 1];
        prefix[i] = prefix[i - 1];
        merge(prefix[i], Data(p[ind] + d[ind], p[ind] - d[ind]));
        if(d[ind] > mx){
            mx2 = mx;
            mx = d[ind];
            ind_mx = ind;
        } else if(d[ind] > mx2) mx2 = d[ind];
    }

    ll lo = mx + mx2, hi = p[n - 1] / 2 + 2e9; // (t - d) / 2 = L - t -> 3*t = 2 * L + d, (L - d) / 3
    lo = max(lo, (p[n - 1] - 1000000000) / 3);
    while(lo < hi){
        ll D = (lo + hi) / 2;
        Data d1, d2;
        int cur = 0;
        for(int i : perm2){
            while(cur < n && d[perm1[cur]] + d[i] + p[i] - p[perm1[cur]] > D) cur++;
            Data temp = prefix[cur];
            if(D < 2 * mx && i == ind_mx){
                temp = Data();
                for(int r = 0; r < cur; r++){
                    int _r = perm1[r];
                    if(_r == i) continue;
                    merge(temp, Data(p[_r] + d[_r], p[_r] - d[_r]));
                }
            }
            d1.a = max(d1.a, temp.a + d[i] + c - p[i] - D);
            d1.b = min(d1.b, temp.b - d[i] - c - p[i] + D);
            d2.a = max(d2.a, temp.a + p[i] - D + d[i] + c);
            d2.b = min(d2.b, temp.b + p[i] + D - d[i] - c);
        }
        
        ll L1 = d1.a, R1 = d1.b;
        ll L2 = d2.a, R2 = d2.b;
        if(L1 > R1 || L2 > R2){
            lo = D + 1;
            continue;
        } 

        bool ok = false;

        int ptr1 = 0;
        int ptr2 = n;
        for(int v = 1; v < n; v++){
            while(ptr1 < n && p[ptr1] - p[v] < L1) ptr1++;
            while(ptr2 > 0 && p[ptr2 - 1] + p[v] >= L2) ptr2--;
            int u = max(ptr1, ptr2);
            if(u >= v) continue;
            if(p[u] - p[v] > R1 || p[u] + p[v] > R2) continue;
            ok = true;
            break;
        }
        if(ok) hi = D;
        else lo = D + 1;
    }
    return lo;
}

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