2020 BIT冬训-二分三分快速幂矩阵 E - Aggressive cows POJ - 2456

Problem 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.

 

 

思路：

将坐标升序排序后。二分查找间隔即可。可利用Lower_bound()函数查找第一个大于等于该值的值所对应的下标（若找不到则返回n（越界））

注意下二分的边界条件（玄学，随机应变）

#include<cstdio>
#include<algorithm>
using namespace std;
int n,c,l,r,mid,p[100005],cnt,temp,flag;
int main(){
    scanf("%d%d",&n,&c);
    for(int i=0;i<n;i++)
        scanf("%d",&p[i]);
    sort(p,p+n);
    l=1,r=p[n-1];
    while(l<=r){
        mid=(l+r)/2;
        flag=0;
        cnt=p[0];
        for(int i=0;i<c-1;i++){
            temp=lower_bound(p,p+n,cnt+mid)-&p[0];
            if(temp==n){
                break;
            }else{
                cnt=p[temp];
            }
            if(i==c-2)
                flag=1;
        }
        if(flag)
            l=mid+1;
        else
            r=mid-1;
    }
    printf("%d",l-1);
}