POJ 2456 Aggressive cows(二分答案)

Posted 2021-02-05 shenben

Aggressive cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 22674		Accepted: 10636

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

 

【题意】

农夫 John 建造了一座很长的畜栏，它包括N (2 <= N <= 100,000)个隔间，这些小隔间依次编号为x1,...,xN (0 <= xi <= 1,000,000,000). 但是，John的C (2 <= C <= N)头牛们并不喜欢这种布局，而且几头牛放在一个隔间里，他们就要发生争斗。为了不让牛互相伤害。John决定自己给牛分配隔间，使任意两头牛之间的最小距离尽可能的大，那么，这个最大的最小距离是什么呢

 

【分析】

贪心+二分
二分枚举相邻两牛的间距，判断大于等于此间距下能否放进所有的牛。 

 

【代码】

#include<cstdio>
#include<algorithm>
#include<iostream>
#define debug(x) cerr<<#x<<" "<<x<<‘
‘;
using namespace std;
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
const int N=1e5+5;
int ans,n,m,d[N];
inline bool check(int now){
	int res=1,p=d[1];
	for(int i=2;i<=n;i++)
		if(d[i]-p>=now)
			res++,p=d[i];
	return res>=m;
}
int main(){
	n=read(),m=read();
	for(int i=1;i<=n;i++) d[i]=read();
	sort(d+1,d+n+1);
	int l=0,r=d[n],mid;
	while(l<=r){
		mid=l+r>>1;
		if(check(mid)){
			ans=mid;
			l=mid+1;
		}
		else{
			r=mid-1;
		}
	}
	printf("%d
",ans);
	return 0;
}