Submission #784945

#	Time	Username	Problem	Language	Result	Execution time	Memory
784945		GusterGoose27	Shortcut (IOI16_shortcut)	C++17	0 / 100	1 ms	212 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) {
    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];
    for (int i = 0; i < n; i++) {
        elems1[i] = pli(pre[i]-d[i], i);
        elems2[i] = pli(pre[i]+d[i], i);
    }
    sort(elems1, elems1+n, greater<pli>());
    sort(elems2, elems2+n, greater<pli>());
    // max_dist[n-1] = d[n-1];
    // int split = n-1;
    // ll rdist = max_dist[n-1];
    // ll ldist = 0;
    // for (int i = n-2; i >= 0; i--) {
    //     ldist = max(ldist, d[i]+sum(i, split));
    //     while (split && ldist >= rdist+l[split-1]) {
    //         split--;
    //         ldist -= l[split];
    //         rdist += l[split];
    //         rdist = max(rdist, (ll)d[split]);
    //     }
    //     max_dist[i] = max(ldist, rdist);
    // }
    ll mn = 0;
    ll mx = inf;
    makeable(150, d, c);
    while (mx > mn+1) {
        ll cur = (mn+mx)/2;
        if (makeable(cur, d, c)) mx = cur;
        else mn = cur;
    }
    return mx;
}

• 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