openjudge-2456——Aggressive cows（二分法）

最新推荐文章于 2021-10-28 16:24:57 发布

星辰浩宇

最新推荐文章于 2021-10-28 16:24:57 发布

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分类专栏：二分

本文链接：https://blog.youkuaiyun.com/amf12345/article/details/97377372

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本文介绍了一个算法问题，即如何在给定的牛棚位置中放置一定数量的牛，使得任意两头牛之间的最小距离尽可能大。通过二分查找和贪心策略，该算法有效地解决了问题，并提供了完整的C++代码实现。

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2456:Aggressive cows

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

样例输出

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.

#include<bits/stdc++.h>
using namespace std;
const int N=100050;
int dst[N];
int n,c;
int main(){
	scanf("%d%d",&n,&c);
	for(int i=1;i<=n;i++){
		scanf("%d",&dst[i]);
	}
	sort(dst+1,dst+n+1);
	int max_x=dst[n]-dst[1];
	int min_x=0;
	while(min_x<max_x){
		int mid=(max_x+min_x+1)/2;
		//每次第一个位置都需要放才能保证最大距离的最小值
		int pre=1,cnt=c-1; 
		for(int i=2;i<=n;i++){
			if(dst[i]-dst[pre]>=mid){
				cnt--;
				pre=i;
				
			}
			if(cnt==0) break;	
		}
		if(cnt==0){
			min_x=mid;
		}else{
			max_x=mid-1;
		}
	}
	cout<<max_x<<endl;
	return 0;
}