Aggressive cows

Aggressive cows

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总时间限制: 

1000ms

 

内存限制: 

65536kB

描述

Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,...,xN (0 <= xi <= 1,000,000,000).

His C (2 <= C <= N) cows don't like this barn layout and become aggressive towards each other once put into a stall. To prevent the cows from hurting each other, FJ want to assign the cows to the stalls, such that the minimum distance between any two of them is as large as possible. What is the largest minimum distance?

输入

* Line 1: Two space-separated integers: N and C

* Lines 2..N+1: Line i+1 contains an integer stall location, xi

输出

* Line 1: One integer: the largest minimum distance

样例输入

5 3
1
2
8
4
9

样例输出

3

提示

OUTPUT DETAILS:

FJ can put his 3 cows in the stalls at positions 1, 4 and 8, resulting in a minimum distance of 3.

Huge input data,scanf is recommended.

#include <iostream>
using namespace std;

int n, c;
int find(int d, int p[]) {
    int t = p[0], count = 1;
    for (int i = 1; i < n; i++) {
        if ((p[i] - t) >= d) {
            t = p[i];
            count++;
        }


    }
    if (count >= c)
        return 1;
    else
        return 0;

}

int cows(int p[]) {
    if (c == 2)
        return p[n - 1] - p[0];
    int i = 0;
    int ans;
    
    int l = 0, r = 2 * p[n - 1] / (c - 1);




    while (l <= r) {

        int  d = (l + r) / 2;


        if (find(d, p)) {
            ans = d;
            l = d + 1;


        } else {
            r = d - 1;

        }



    }
    return ans;
}


void merge_sort(int arr[], int len) {
    int *a = arr;
    int *b = (int *) malloc(len * sizeof(int));
    int seg, start;
    for (seg = 1; seg < len; seg += seg) {
        for (start = 0; start < len; start += seg + seg) {
            int low = start, mid = min(start + seg, len), high = min(start + seg + seg, len);
            int k = low;
            int start1 = low, end1 = mid;
            int start2 = mid, end2 = high;
            while (start1 < end1 && start2 < end2)
                b[k++] = a[start1] < a[start2] ? a[start1++] : a[start2++];
            while (start1 < end1)
                b[k++] = a[start1++];
            while (start2 < end2)
                b[k++] = a[start2++];
        }
        int *temp = a;
        a = b;
        b = temp;
    }

    if (a != arr) {
        int i;
        for (i = 0; i < len; i++) {
            //cout << b[i] << endl;
            b[i] = a[i];
        }
        b = a;

    }
    free(b);
}

int main() {


    cin >> n >> c;
    int p[n];

    for (int i = 0; i < n; i++) {
        cin >> p[i];
    }
    merge_sort(p, n);
    cout << cows(p);
    return 0;
}