POJ 2456 Aggressive cows<二分贪心>

Aggressive cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11952		Accepted: 5842

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.

二分的题目，用二分加判断函数，判断函数里，我们只要对二分传进来的距离进行贪心，看这一种的距离下是否能放入C头牛，能放入的话就代表可行，继续向后寻找更大的可能

#include<cstdio>
#include<cctype>
#include<iostream>
#include<stack>
#include<map>
#include<cstring>
#include<string>
#include<sstream>
#include<queue>
#include<set>
#include<algorithm>
using namespace std ;
int a[100005];
int n,c;
int f(int mid){
int ans=1;
int tp=a[0];
for(int i=1;i<n;i++){
if(a[i]-tp>=mid)
{
ans++;
tp=a[i];
}
if(ans>=c)
return 1;
}

return 0;
}
int main()
{
while(~scanf("%d %d",&n,&c)){
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
sort(a,a+n);
int l=1,r=a[n-1];
while(r-l>1){
int m=(r+l)/2;
if(f(m)) l=m;
else
r=m;
}
printf("%d\n",l);
}
return 0 ;
}