USACO 2006 DEC #3(Silver): River Hopscotch

Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at the end, L units away from the start (1 <= L <= 1,000,000,000). Along the river between the starting and ending rocks, N (0 <= N <= 50,000) more rocks appear, each at an integral distance Di from the start (0 < Di < L).

To play the game, each cow in turn starts at the starting rock and tries to reach the finish at the ending rock, jumping only from rock to rock. Of course, less agile cows never make it to the final rock, ending up instead in the river.

Farmer John is proud of his cows and watches this event each year. But as time goes by, he tires of watching the timid cows of the other farmers limp across the short distances between rocks placed too closely together. He plans to remove several rocks in order to increase the shortest distance a cow will have to jump to reach the end. He knows he cannot remove the starting and ending rocks, but he calculates that he has enough resources to remove up to M rocks (0 <= M <= N).

FJ wants to know exactly how much he can increase the shortest distance *before* he starts removing the rocks. Help Farmer John determine the greatest possible shortest distance a cow has to jump after removing the optimal set of M rocks.

INPUT FORMAT:

Line 1: Three space-separated integers: L, N, and M

Lines 2..N+1: Each line contains a single integer indicating how far some rock is away from the starting rock. No two rocks share the same position.

OUTPUT FORMAT:

Line 1: A single integer that is the maximum of the shortest distance a cow has to jump after removing M rocks

SAMPLE INPUT:

25 5 2
2
14
11
21
17

SAMPLE OUTPUT:

4

This is also a binary search question, if you think about it; if you know that every rocks are at least d distance apart after removing some rocks, then if the amount of rocks you remove are smaller than M, then the maximum d you can obtain would be the answer for this question.

Here’s my code:

//
//  main.cpp
//  제자리 멀리뛰기
//
//  Created by Shin Lee on 24/03/2020.
//  Copyright © 2020 Shin Lee. All rights reserved.
//

#include <bits/stdc++.h>
using namespace std;
int n, d, m; int arr[50005];int dist[50005];
int main(){
    cin >> d >> n >> m;
    for(int i=0;i<n;i++){
        cin >> arr[i];
    }
    arr[n]=0; arr[n+1]=d;
    sort(arr,arr+n+2);
    
    int hi = d, lo = 1;
    while(hi>lo){
       // printf("hi = %d, lo = %d\n",hi,lo);
        int mid = (hi+lo+1)/2;
        int pos=0;int cnt=0;
        for(int i=1;i<=n+1;i++){
            if(arr[i]-arr[pos]<mid){

                cnt++;
            }else {
                pos = i;
            }
        }
        if(cnt<=m){
            lo = mid;
        }else hi = mid-1;
        
    }
    cout << lo;
    
}