Submission #139310

#	Time	Username	Problem	Language	Result	Execution time	Memory
139310		BThero	Shortcut (IOI16_shortcut)	C++17	71 / 100	2068 ms	318176 KiB

shortcut

// Why am I so dumb? :c
// chrono::system_clock::now().time_since_epoch().count()

//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include "shortcut.h"

#define pb push_back
#define mp make_pair

#define all(x) (x).begin(), (x).end()
#define debug(x) cerr << #x << " = " << x << '\n';

#define fi first
#define se second

using namespace std;
using namespace __gnu_pbds;

typedef long long ll;   
template<typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

const int MAXN = (int)1e5 + 5;
const int MAXP = (int)1e7 + 5;
const ll INF = (ll)1e17;

struct Rect {
    ll lx, rx, ly, ry;
    
    Rect() {
        lx = ly = -INF;
        rx = ry = INF;
    }
    
    Rect(ll lx, ll rx, ll ly, ll ry)
        : lx(lx), rx(rx), ly(ly), ry(ry) {}
    
    void operator = (Rect b) {
        lx = b.lx;
        ly = b.ly;
        rx = b.rx;
        ry = b.ry;
    }  
};

Rect combine(Rect a, Rect b) {
    return Rect(max(a.lx, b.lx), min(a.rx, b.rx), max(a.ly, b.ly), min(a.ry, b.ry));
}

ll w[MAXN], d[MAXN], p[MAXN];

int c1[MAXP], c2[MAXP];

Rect T[MAXP];

int root, sz;

int n;

ll k;

int newVer() {
    T[sz] = Rect();
    c1[sz] = c2[sz] = -1;
    return sz++;
}

void update(int &v, ll tl, ll tr, ll p, Rect x) {
    if (v == -1) {
        v = newVer();
    }
    
    if (tl == tr) {
        T[v] = combine(T[v], x);
        return;
    }
    
    ll mid = (tl + tr) >> 1;
    
    if (p <= mid) {
        update(c1[v], tl, mid, p, x);
    }
    else {
        update(c2[v], mid + 1, tr, p, x);
    }
    
    T[v] = Rect();
    
    if (c1[v] != -1) {
        T[v] = combine(T[v], T[c1[v]]);
    }
    
    if (c2[v] != -1) {
        T[v] = combine(T[v], T[c2[v]]);
    }
}

Rect query(int &v, ll tl, ll tr, ll l, ll r) {
    if (l > r || tl > r || tr < l || v == -1) {
        return Rect();
    }
    
    if (l <= tl && tr <= r) {
        return T[v];
    }
    
    ll mid = (tl + tr) >> 1;
    return combine(query(c1[v], tl, mid, l, r), query(c2[v], mid + 1, tr, l, r));
}

bool check(ll lim) {
    ll tax = -INF, tbx = INF;
    ll tay = -INF, tby = INF;
    sz = 0;
    root = -1;
    
    for (int i = n; i > 0; --i) {
        Rect cur = query(root, -INF, INF, lim + p[i - 1] - d[i] + 1, INF);
        tax = max(tax, p[i - 1] + d[i] - lim + k + cur.lx);
        tbx = min(tbx, p[i - 1] - d[i] + lim - k + cur.rx);
        tay = max(tay, p[i - 1] + d[i] - lim + k + cur.ly);
        tby = min(tby, p[i - 1] - d[i] + lim - k + cur.ry);
        
        cur.lx = p[i - 1] + d[i];
        cur.rx = p[i - 1] - d[i];
        cur.ly = -p[i - 1] + d[i];
        cur.ry = -p[i - 1] - d[i];
        
        update(root, -INF, INF, d[i] + p[i - 1], cur);
    }

    //for (int a = 1; a <= n; ++a) {
        //for (int b = a + 1; b <= n; ++b) {
            //if (d[a] + d[b] + p[b - 1] - p[a - 1] <= lim) {
                //continue;
            //}
            
            //ll cx = p[a - 1] + p[b - 1];
            //ll cy = p[a - 1] - p[b - 1];
            //ll r = lim - k - d[a] - d[b];
            
            //if (r < 0) {
                //return 0;
            //}
            
            //ll ax = cx - r, bx = cx + r;
            //ll ay = cy - r, by = cy + r;
            
            //tax = max(tax, ax);
            //tbx = min(tbx, bx);
            //tay = max(tay, ay);
            //tby = min(tby, by);
        //}
    //}
    
    for (int b = n; b > 1; --b) {
        ll l = tax - p[b - 1];
        ll r = tbx - p[b - 1];
        l = max(l, tay + p[b - 1]);
        r = min(r, tby + p[b - 1]);
        
        int pos = lower_bound(p, p + n, l) - p + 1;
        
        if (pos < b && p[pos - 1] <= r) {
            return 1;
        }
    }
    
    return 0;
}

ll solve() {
    for (int i = 1; i < n; ++i) {
        p[i] = p[i - 1] + w[i];
    }
    
    ll l = 0, r = INF, ans = -1;
    
    while (l <= r) {
        ll mid = (l + r) >> 1;
        
        if (check(mid)) {
            ans = mid;
            r = mid - 1;
        }
        else {
            l = mid + 1;
        }
    }
    
    return ans;
}

ll find_shortcut(int _n, vector<int> l, vector<int> _d, int c) {
    n = _n;
    k = c;
    
    for (int i = 1; i < n; ++i) {
        w[i] = l[i - 1];
    }
    
    for (int i = 1; i <= n; ++i) {
        d[i] = _d[i - 1];
    }

    return solve();
}

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