Submission #1116764

#	Time	Username	Problem	Language	Result	Execution time	Memory
1116764		SalihSahin	Shortcut (IOI16_shortcut)	C++14	31 / 100	2069 ms	508 KiB

shortcut

#include <bits/stdc++.h>
#define pb push_back
using namespace std;
#include "shortcut.h"

long long inf = 1e16;

long long find_shortcut(int n, vector<int> l, vector<int> d, int c){
    vector<long long> pre(n);
    for(int i = 0; i < n-1; i++){
        pre[i+1] = pre[i] + l[i];
    }
    vector<long long> lfar(n), rfar(n);
    lfar[0] = d[0];
    for(int i = 1; i < n; i++){
        if(d[i] > lfar[i-1] + l[i-1]) lfar[i] = d[i];
        else lfar[i] = lfar[i-1] + l[i-1];
    }
    rfar[n-1] = d[n-1];
    for(int i = n-2; i >= 0; i--){
        if(d[i] > rfar[i+1] + l[i]) rfar[i] = d[i];
        else rfar[i] = rfar[i+1] + l[i];
    }

    vector<long long> precalcl(n), precalcr(n);
    for(int i = 1; i < n; i++){
        precalcl[i] = max(precalcl[i-1], (long long)(lfar[i-1] + l[i-1] + d[i]));
    }
    for(int i = n-2; i >= 0; i--){
        precalcr[i] = max(precalcr[i+1], (long long)(rfar[i+1] + l[i] + d[i]));
    }

    long long ans = inf;
    bool calc = false;

    for(int l = 0; l < n; l++){
        for(int r = l+1; r < n; r++){
            if(pre[r] - pre[l] < c && calc) continue;

            calc = true;
            long long dis = lfar[l] + rfar[r] + min((long long)c, pre[r] - pre[l]); // L-R
            dis = max(dis, precalcl[l]); // L-L
            dis = max(dis, precalcr[r]); // R-R
            long long clen = pre[r] - pre[l] + c;
            
            for(int j = l+1; j <= r; j++){
                dis = max(dis, lfar[l] + (long long)d[j] + min(pre[j] - pre[l], clen - (pre[j] - pre[l])));
            } // L-M
            for(int j = l; j < r; j++){
                dis = max(dis, rfar[r] + (long long)d[j] + min(pre[r] - pre[j], clen - (pre[r] - pre[j])));
            } // R-M

            for(int j = l+1; j < r; j++){ // M-M
                long long addj = d[j];
                for(int j2 = j+1; j2 < r; j2++){
                    long long addj2 = d[j2];
                    dis = max(dis, min(pre[j2] - pre[j], clen - (pre[j2] - pre[j])) + addj + addj2);
                }
            }

            ans = min(ans, dis);
        }
    }

    return ans;
}

• 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