Submission #1116868

#	Time	Username	Problem	Language	Result	Execution time	Memory
1116868		SalihSahin	Shortcut (IOI16_shortcut)	C++14	38 / 100	2058 ms	760 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;
    long long val = 0;
    for(int i = 0; i < n-1; i++){
        val = max(val, lfar[i] + rfar[i+1] + l[i]);
    }
    ans = val;
 
    for(int l = 0; l < n-1; l++){
        long long maxla = precalcl[l];
 
        int l2 = l+1, r2 = n-1;
        while(l2 < r2){
            int r = (l2 + r2)/2;
            long long dis1 = 0;
            dis1 = lfar[l] + rfar[r] + c; // L-R
            dis1 = max(dis1, precalcr[r]); // R-R
            long long clen = pre[r] - pre[l] + c;
            for(int j = l; j < r; j++){
                dis1 = max(dis1, rfar[r] + (long long)d[j] + min(pre[r] - pre[j], clen - (pre[r] - pre[j])));
            } // R-M

            long long dis2 = 0;
            
            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+1; j < r; j++){ // M-M
                for(int j2 = j+1; j2 < r; j2++){
                    dis2 = max(dis2, min(pre[j2] - pre[j], clen - (pre[j2] - pre[j])) + (long long)d[j] + (long long)d[j2]);
                }
            }
            ans = min(ans, max({dis1, dis2, maxla}));
            if(dis1 < dis2) r2 = r;
            else l2 = r + 1;
        }

        int r = l2;

        long long dis1 = 0;
        dis1 = lfar[l] + rfar[r] + c; // L-R
        dis1 = max(dis1, precalcr[r]); // R-R
        long long clen = pre[r] - pre[l] + c;
        for(int j = l; j < r; j++){
            dis1 = max(dis1, rfar[r] + (long long)d[j] + min(pre[r] - pre[j], clen - (pre[r] - pre[j])));
        } // R-M

        long long dis2 = 0;
        
        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+1; j < r; j++){ // M-M
            for(int j2 = j+1; j2 < r; j2++){
                dis2 = max(dis2, min(pre[j2] - pre[j], clen - (pre[j2] - pre[j])) + (long long)d[j] + (long long)d[j2]);
            }
        }
 
        ans = min(ans, max({dis1, dis2, maxla}));
    }
 
    return ans;
}

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