Submission #784948

#	Time	Username	Problem	Language	Result	Execution time	Memory
784948		GusterGoose27	Shortcut (IOI16_shortcut)	C++17	23 / 100	2078 ms	340 KiB

shortcut

#include "shortcut.h"

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<ll, int> pli;
typedef pair<ll, ll> pll;

const int MAXN = 1e5+5;
const int MAXM = 1e9+5;
const ll inf = (ll)MAXN*MAXM;
ll pre[MAXN];
ll max_dist[MAXN];
pli elems1[MAXN];
pli elems2[MAXN];
int n;

ll sum(int l, int r) {
    if (l > r) swap(l, r);
    return pre[r]-pre[l];
}

pll max(pll a, ll b) {
    if (b > a.first) return pll(b, a.first);
    if (b > a.second) return pll(a.first, b);
    return pll(a.first, a.second);
}

bool makeable(ll cur, vector<int> d, int c) {
    ll mx1 = -inf;
    ll mx2 = -inf;
    int j = 0;
    // priority_queue<int> left;
    // priority_queue<int, vector<int>, greater<int>> right;
    pll lval(-inf, -inf);
    pll rval(-inf, -inf);
    ll sum = 0;
    for (int i = 0; i < n; i++) {
        while (j < n && elems2[j].first > cur+elems1[i].first) {
            int v = elems2[j].second;
            lval = max(lval, pre[v]+d[v]);
            rval = max(rval, d[v]-pre[v]);
            j++;
        }
        int v = elems1[i].second;
        ll lv = (lval.first == pre[v]+d[v]) ? lval.second : lval.first;
        ll rv = (rval.first == d[v]-pre[v]) ? rval.second : rval.first;
        int mn = -1; // max val leq this
        int mx = n;
        while (mx > mn+1) {
            int split = (mn+mx)/2;
            if (2*pre[split] <= lv-rv) mn = split;
            else mx = split;
        }
        sum = inf;
        if (mn >= 0) sum = lv-pre[mn];
        if (mx < n) sum = min(sum, rv+pre[mx]); 
        mx1 = max(mx1, c+d[v]-pre[v]+sum);
        mx2 = max(mx2, c+d[v]+pre[v]+sum);
        // assert(mx2 < 140);
    }
    for (int i = 0; i < n; i++) {
        if (cur >= mx1+pre[i] && cur >= mx2-pre[i]) return 1;
    }
    return 0;
}

ll find_shortcut(int N, vector<int> l, vector<int> d, int c) {
    n = N;
    for (int i = 1; i < n; i++) pre[i] = pre[i-1]+l[i-1];
    ll bval = inf;
    for (int u = 0; u < n; u++) {
        for (int v = u+1; v < n; v++) {
            ll cur_val = 0;
            for (int a = 0; a < n; a++) {
                for (int b = a+1; b < n; b++) {
                    cur_val = max(cur_val, d[a]+d[b]+min(sum(a, b), c+sum(a, u)+sum(b, v)));
                }
            }
            bval = min(bval, cur_val);
        }
    }
    return bval;
}

• Subtask 1: 9 / 9
• Subtask 2: 14 / 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