二分法的应用：最大化最小值 POJ2456 Aggressive cows

最新推荐文章于 2025-09-07 20:32:33 发布

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最新推荐文章于 2025-09-07 20:32:33 发布

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文章标签： java

原文链接：http://www.cnblogs.com/Alpharun/p/8663510.html

本文介绍了一个使用二分法解决最大化最小距离问题的经典案例：Aggressive Cows（好斗的奶牛）。该问题的目标是在给定的牛棚位置中放置一定数量的奶牛，使得任意两只奶牛之间的最小距离尽可能大。文章通过具体实例演示了如何实现这一算法，并给出了完整的Java代码。

/*
二分法的应用：最大化最小值  POJ2456 Aggressive cows

Time Limit: 1000MS        Memory Limit: 65536K
Total Submissions: 18125        Accepted: 8636
Description

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 Input

5 3
1
2
8
4
9
Sample Output

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.
Source

USACO 2005 February Gold
 */

import java.util.Arrays;
import java.util.Scanner;

public class Main {
    static int N, M;
    static int[] a;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        N = sc.nextInt();
        M = sc.nextInt();
        a = new int[N];
        for (int i = 0; i < N; i++)
            a[i] = sc.nextInt();
        sc.close();
        Arrays.sort(a);
        int low = -1, high = 1000000001;
        while (high - low > 1) {
            int mid = (low + high) / 2;
            if (C(mid)) {
                low = mid;
            } else {
                high = mid;
            }
        }
        System.out.println(low);
    }

    static boolean C(int X) {
        int last = 0;
        for (int i = 1; i < M; i++) {
            int next = last + 1;
            while (next < N && a[next] - a[last] < X) {
                next++;
            }
            if (next >= N) return false;
            last = next;
        }
        return true;
    }
}