POJ 2456 Aggressive cows (二分 基础)

Aggressive cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7924		Accepted: 3959

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： http://poj.org/problem?id=2456

题目大意：一个数轴上n个点，每个点一个整数值，有c个物品，要放在这些点的某几个上，求怎么放可以使任意两个物品间距离的最小值最大，求这个最大值

题目分析：最小值最大，典型二分题，二分距离的值判断

#include 
#include 
using namespace std;
int const INF = 0x3fffffff;
int const MAX = 1e5 + 5;
int d[MAX];
int n, c;

int cal(int m)
{	
	int ans = 0, now = d[0];
	for(int i = 1; i < n; )
	{
		while(d[i] < now + m)
			i ++;
		ans ++;
		now = d[i];
	}
	return ans;
}	

int main()
{
	scanf("%d %d", &n, &c);
	for(int i = 0; i < n; i++)
		scanf("%d", &d[i]);
	sort(d, d + n);
	int r = d[n - 1] - d[0], l = 0, ans = 0;
	while(l <= r)
	{
		int m = (l + r) / 2;
		int num = cal(m);
		if(num >= c)
		{
			ans = m;
			l = m + 1;
		}
		else
			r = m - 1;
	}
	printf("%d\n", ans);
}