Aggressive cows 二分不仅仅是查找

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7083		Accepted: 3522

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.

Source

USACO 2005 February Gold

 

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <string>
 #include <vector>
 #include <set>
 #include <map>
 #include <queue>
 #include <stack>
 #include <sstream>
 #include <iomanip>
 using namespace std;
 const int INF=0x4fffffff;
 const int EXP=1e-;
 const int MS=;

 int pos[MS];
 int N,M;

 bool judge(int d)
 {
       int last=;
       int now;
       for(int i=;i<M;i++)     //贪心法给第i头牛找宿舍
       {
             now=last+;
             while(now<N&&pos[now]-pos[last]<d)
                   now++;
             if(now>=N)
                   return false;
             last=now;
       }
       return true;
 }

 void solve()
 {
       sort(pos,pos+N);
       int l=,r=INF;
       while(r-l>)         //  注意是区间,很多时候开区间更方便一些。
       {
             int mid=(l+r)/;
             if(judge(mid))
                   l=mid;
             else
                   r=mid;
       }
       printf("%d\n",l);
 }

 int main()
 {
       while(scanf("%d%d",&N,&M)!=EOF)
       {
             for(int i=;i<N;i++)
                   scanf("%d",&pos[i]);
             solve();
       }
       return ;
 }