POJ - 2456 Aggressive cows 二分 最大化最小值

Aggressive cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18099		Accepted: 8619

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

题目链接：点击打开链接

题目大意：给牛分配隔间，使任意两头牛之间的最小距离尽可能的大

思路：先将牛的位置按从小到大的顺序排序，再二分枚举查找，左值为0，右值为最左端与最右端的牛的距离。

#include<stdio.h>
#include<algorithm>
using namespace std;

int n, c,x[100005];

int check(int mid)
{
	int l = 0;
	for (int i = 1; i < c; i++)
	{
		int r = l + 1;
		while (r < n&&x[r] - x[l] < mid)
			r++;
		if (r == n)return 0;
		l = r;
	}
	return 1;
}


int main()
{
	while (~scanf("%d %d", &n, &c))
	{
		for (int i = 0; i < n; i++)
			scanf("%d", &x[i]);
		sort(x, x + n);
		int left = 0, right = x[n-1];
		while (right > left +1)
		{
			int mid = (right + left) / 2;
			if (check(mid))
				left = mid;
			else
				right = mid;
		}
		printf("%d\n", left);
	}

	return 0;
}