POJ 2456 Aggressive cows__二分-优快云博客

本文链接：https://blog.youkuaiyun.com/tp7309/article/details/54144020

本文介绍了一个经典的算法问题：如何在给定的牛棚位置中放置一定数量的牛，使得任意两头牛之间的最小距离尽可能大。通过使用二分查找的方法，文章详细解释了算法的设计思路与实现过程。

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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.

分析

确定n个不同位置的摊位可以放几头牛，要算最大间隔，一个求最大值的问题，用二分法，确认最大、最小间距后不断寻找即可。

实现

#include <iostream>
#include <cstring>
#include <algorithm>
#define MAXN 100010
int a[MAXN];
int n, c;

bool canSolve(long long ans)
{
    int lastCowStallIndex = 0, cowCount = 1; //第一头牛肯定放第一个摊位。
    for (int i = 1; i < n; i++) {
        if (a[i] - a[lastCowStallIndex] >= ans) { //大于给定间距则说明可以再放一头牛。
            cowCount++;
            lastCowStallIndex = i;
            if (cowCount >= c) return true;
        }
    }
    return false;
}

int main()
{
//  freopen("in.txt", "r", stdin);
    scanf("%d%d", &n, &c);
    for (int i = 0; i < n; i++) {
        scanf("%d", &a[i]);
    }
    if (n <= 0 || c <= 0 || (n < c)) {
        printf("%d\n", 0);
        return 0;
    }
    std::sort(a, a + n);
    long long l = (n - a[0]) / (c - 1);
    l = l < 0 ? 0 : l;
    long long r = (a[n - 1] - a[0] / (c - 1));
    while (l <= r) {
        long long mid = (l + r) >> 1;
        canSolve(mid) ? (l = mid + 1) : (r = mid - 1);
    }
    printf("%lld", l - 1);
    return 0;
}