Aggressive cows

原创已于 2022-03-28 15:44:56 修改 · 568 阅读

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#c++ #算法

于 2022-03-28 15:20:39 首次发布

解决农场中牛棚分配问题，通过二分查找算法确定最大最小间距，确保任意两头牛间的距离尽可能大。

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?

Input

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

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

Output

* Line 1: One integer: the largest minimum distance

Sample

Inputcopy	Outputcopy
5 3 1 2 8 4 9	3

Hint

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.

题目的思想就是二分确定判断间距的范围，然后利用二分去判断距离是否符合要求，并不断进行更新；

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;
const int N = 100010, INF = 0x3f3f3f3f;

int q[N]; //储存位置
int n, m; // 数量， 牛数

bool judge(int k) { // k表示的是最小的距离，两个放牛的棚子之间的距离要大于k
    int sum = 1; // 已经放了牛的数量
    int now = 1; //上一个放牛的棚子的编号
    for (int i = 2; i <= n; i++) {
        if (q[i] - q[now] >= k) {
            now = i;
            sum ++;
        }
    }
    if (sum >= m) return true;
    else return false ;
}

int main() {
    int l = INF, r = -INF;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &q[i]);
        l = min(q[i], l);
        r = max(q[i], r);
    }

    sort(q + 1, q + n + 1);

    int res;
    r = r / m; // 距离最大的情况
    while (l <= r) {
        int mid = l + r >> 1; //mid表示距离
        if (judge(mid) == true) {
            l = mid + 1;
            res = mid;
        } else {    
            r = mid - 1;
        }
    }
    cout << res;
    return 0;
}