Aggressive cows （北京大学ACM-ICPC竞赛训练暑期课 ）

标签：nbsp   details   each   can   name   cat   minimum   eve   位置   

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

5 3
1
2
8
4
9

样例输出

3

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 <iostream>
#include <algorithm>
using namespace std;
int n,k;
int a[100000+5],b[ 100000+5];
int fun(int m)
{
    int cnt=0;
    while(m)
    {
        if(m%2) cnt++;
        m>>=1;
    }
    return cnt;
}
int judge(int m)
{
    int cnt=1,temp=a[0];//第一个位置有牛 
    for(int i=1;i<n;i++)
    {
        if(a[i]-temp>=m) 
        {
            temp=a[i];
            cnt++;
        }
    }
    if(cnt>=k) return 1;
    else return 0;
}
void solve()
{
    int l=0,r=a[n-1]-a[0];//距离 
    while(l<=r)
    {
        int mid=l+(r-l)/2;
        if(judge(mid))  l=mid+1;
        else r=mid-1;
    } 
    cout<<l-1<<endl;
    return ;
}
int main()
{
    while(cin>>n>>k)
    {
        for(int i=0;i<n;i++) cin>>a[i];
        sort(a,a+n);
        solve();
    }
    return 0;
} 

Aggressive cows （北京大学ACM-ICPC竞赛训练暑期课 ）

标签：nbsp   details   each   can   name   cat   minimum   eve   位置   

原文地址：https://www.cnblogs.com/Shallow-dream/p/11553151.html