POJ2456 Aggressive cows(对整数二分)

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农夫约翰新建了一座长形牛棚，拥有N个牛位，分布于直线上。由于牛群间存在攻击倾向，约翰需要合理分配牛位，确保牛之间的最小距离尽可能大。本文提供了解决该问题的算法实现，包括输入格式、输出说明及样例输入输出。通过二分查找算法，有效解决了牛棚布局问题。

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

Sample Output

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.

二分位置

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
int a[100005], n, c;
bool judge(int x)
{
	int cnt = 1, tmp = a[0];
	for(int i = 0; i < n; ++i)
		if(a[i] - tmp >= x) {
			cnt++;
			tmp = a[i];
			if(cnt >= c) return true;
		}
	return false;
}
int serch()
{
	int l = 0, r = a[n - 1] - a[0];
	while(l <= r) {
		int mid = (l + r) / 2;
		if(judge(mid)) l = mid + 1;
		else r = mid - 1;
	}
	return l - 1;
}
int main(int argc, char const *argv[])
{
	while(scanf("%d%d", &n, &c) != EOF) {
		for(int i = 0; i < n; ++i)
			scanf("%d", &a[i]);
		sort(a, a + n);
		printf("%d\n", serch());
	}
	return 0;
}