E - 05_e-05-优快云博客

本文链接：https://blog.youkuaiyun.com/lml11111/article/details/70767786

本文解决了一个经典的二分搜索问题——如何将一定数量的奶牛分配到沿直线分布的多个谷仓中，使得任意两个奶牛之间的最小距离尽可能大。通过使用C语言实现的二分搜索算法，文章详细介绍了其解决方案及代码实现。

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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

Sample Output

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.

题意：
谷仓的位置在一条线上，把谷仓分成c份，使谷仓的最小距离最大

分析：
就是用二分

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int a[100005],c,n;
int vp(long int g)
{
    int i,gr,sum;
    gr=a[0];
    sum=1;
    for(i=1;i<n;i++)
    if(a[i]-gr>=g)
    {
       gr=a[i];
       sum++;
    }
    return sum;
}
int main()
{
    int i,l,k,mid;
    scanf("%d%d",&n,&c);
    for(i=0;i<n;i++)
        scanf("%d",&a[i]);
    sort(a,a+n);
    l=0;
    k=a[n-1]-a[0];
    while(l<=k)
    {
        mid=(k+l)/2;
        if(vp(mid)>=c)l=mid+1;
        else
        if(vp(mid)<c)
            k=mid-1;
    }
    printf("%d\n",l-1);
}

我发现用c++老是超时，必须改成c,主要是c不熟啊