POJ2456 Aggressive cows

Aggressive cows

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11665 Accepted: 5716

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 312849

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：求最短间隔的最大值。
解题思路：典型的二分查找题目，用二分查找猜测可能的最大间距即可。对于每一个猜测值，看看在这个距离下能容纳多少牛，如果可容纳的牛数多余总牛数，说明距离过短，扩大左边界；否则，距离过长，减小右边界。

#include <iostream>#include <cstdio>#include <stdlib.h>using namespace std;int pos[100005];int N,C;int cmp(const void* a,const void* b)  //比较函数{return *(int*)a-*(int*)b;}bool check(int mid){int cnt=0,pre=pos[0];for(int i=1;i<N&&cnt<C;i++){if(pos[i]-pre>=mid){pre=pos[i];cnt++; //cnt计数在mid的间距下能容纳的总牛数}}if(cnt>=C-1) return true; //C头牛有C-1个间距return false;}int main(){    scanf("%d%d",&N,&C);    for(int i=0;i<N;i++){scanf("%d",&pos[i]);    }    qsort(pos,N,sizeof(int),cmp);//快速排序    int l,r,mid;    //设置左右边界    l=0;    r=pos[N-1];    mid=l+(r-l)/2;//设置中间值，相当于（l+r）/2 ，这样写防止（l+r）溢出    while(l+1<r){ //二分查找最短间隔的最大值mid=l+(r-l)/2;if(check(mid)) l=mid;else r=mid;    }    printf("%d",l);    return 0;}