二分搜索——Aggressive cows

最新推荐文章于 2021-01-02 14:38:33 发布

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本文探讨了一个经典的算法问题：如何在给定的牛棚位置中，安排一定数量的牛，使得任意两头牛之间的最小距离尽可能大。通过使用二分查找和贪心策略，我们能够有效地找到最优解。

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									Aggressive cows

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.

#include<iostream>
#include<cstdlib>
#include<string.h>
#include<vector>
#include<algorithm>
using namespace std;
long long a[100010];
int n,c;
int check(long long mid){
	int cnt,i,way;
	cnt = 1;
	way = a[0];
	for(i = 1;i<n;i++){
		if(a[i]-way>=mid){
			cnt++;
			way = a[i];
		}
	}
	if(cnt>=c){
			return 1;
	}
	return 0;
}
int main(){
	int i;
	long long left,right,ans,mid;
	scanf("%d %d",&n,&c);
		for(i = 0;i<n;i++){
			scanf("%lld",&a[i]);
		}
		sort(a,a+n);
		left = 1;right = a[n-1]-a[0];ans = 0;
		while(left <= right){
			mid = (left+right)/2;
			if(check(mid)){
				ans = mid;
				left = mid+1;
			}
			else{
				right = mid-1;
			}
		}
		printf("%lld\n",ans);
	
}