Submission #1116774

#	Time	Username	Problem	Language	Result	Execution time	Memory
1116774		SalihSahin	Shortcut (IOI16_shortcut)	C++14	0 / 100	1 ms	336 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;

    for(int l = 0; l < n; l++){
        long long maxla = precalcl[l];

        int r;
        int L = l+1, R = n-1;
        while(L < R){
            int M = (L + R)/2;
            if(M >= n && M <= l) break;
            r = M;

            long long dis1 = 0, dis2 = 0;
            dis1 = max(dis1, precalcr[r]); // R-R
            dis1 = lfar[l] + rfar[r] + min((long long)c, pre[r] - pre[l]); // L-R

            long long clen = pre[r] - pre[l] + c;
            
            for(int j = l+1; j <= r; j++){
                dis2 = max(dis2, 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++){
                dis2 = max(dis2, 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];
                    dis2 = max(dis2, min(pre[j2] - pre[j], clen - (pre[j2] - pre[j])) + addj + addj2);
                }
            }

            ans = min(ans, max({maxla, dis1, dis2}));
            if(dis1 >= dis2) L = M + 1;
            else R = M;
        }

        if(L >= n && L <= l) continue;

        r = L;
        long long dis1 = 0, dis2 = 0;
        dis1 = max(dis1, precalcr[r]); // R-R
        dis1 = lfar[l] + rfar[r] + min((long long)c, pre[r] - pre[l]); // L-R

        long long clen = pre[r] - pre[l] + c;
        
        for(int j = l+1; j <= r; j++){
            dis2 = max(dis2, 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++){
            dis2 = max(dis2, 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];
                dis2 = max(dis2, min(pre[j2] - pre[j], clen - (pre[j2] - pre[j])) + addj + addj2);
            }
        }

        ans = min(ans, max({maxla, dis1, dis2}));
    }

    return ans;
}

• 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