C语言数轴上数的最近邻距离的最大值二分法实现

1: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.

题目求这个函数的极值，思路是直接枚举这个极大值d，假设这个最大值是d，那么寻找C个满足这个值的数组，直到找到目标。这里是采用依次寻找的方法，即第一个位置就放在x[0]，依次放入后面的位置，如果满足的解的个数大于C，说明这个距离不够大，若否则说明这个d值取得最大。

所以程序最外层while 是个标准的二分程序，上面方的过程一个for循环就能实现

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
#define maxn 100010
int x[maxn];

int main()
{
	int n, c;
	scanf_s("%d %d", &n, &c);
	for (int i = 0; i < n; i++)
	{
		scanf_s("%d", &x[i]);
	}
	sort(x, x + n);
	int l = x[0];
	int r = x[n - 1];
	while (l<r)
	{
		int last = x[0];
		int cnt = 1;
		int mid = (l + r) / 2;
		for (int i = 0; i<n; i++)
		{
			if (x[i] - last >= mid) 
			{
				cnt++;
				last = x[i];
			}
		}
		if (cnt >= c)
			l = mid + 1;//在此最小距离下足以系下c头牛，说明可以尝试继续增大
		else
			r = mid;
	}
	printf("%d\n", l-1);
	return 0;
}