Aggressive cows （二分）

Aggressive cows

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 37   Accepted Submission(s) : 13

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<br><br>* 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 312849

 

Sample Output

3

 

Source

PKU

 

题意：给你n个坐标，求c个坐标中最小的最大距离。

思路：

一看题目是最小值得最大值，立刻想到二分。对坐标进行排序。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#include <iomanip>#define maxn 15#define mod 1000000007#define INF 0x3f3f3f3f#define exp 1e-6#define pi acos(-1.0)using namespace std;int a[100005];int main(){    ios::sync_with_stdio(false);  //不加这句话会超时    int n,c;    int i,j;    cin>>n>>c;    int m=n-c;  //需要去掉的个数    memset(a,0,sizeof(a));    for(i=1;i<=n;i++)cin>>a[i];    sort(a+1,a+n+1);    int left=0,right=a[n],mid;    while(right>left)    {        int cnt=0;        mid=(left+right)/2;        for(i=2,j=1;i<=n;){   //从第二个开始和第一个进行比较            if(a[i]-a[j]<=mid){cnt++;i++;}            else {j=i;i++;}        }        if(cnt>m) right=mid;          else left=mid+1; //扩大    }    cout<<left<<endl;}