二分贪心-E

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.

这道题大致的意思是有n个牛舍，每个牛舍在自己的坐标xi，把c头牛放在这n个牛舍里面，求最近两头牛之间的最大距离。用二分法判断即可。

下面是我的AC代码：

 

#include<iostream> #include<algorithm> #include<cstdio> using namespace std; int x[100010]; int N,C; bool fun(int mid) { int cnt=1; int m=x[1]; for(int i=2; i<=N; i++) { if(x[i]-m>=mid) { cnt++; m=x[i]; } if(cnt>=C) return true; } return false; } int main() { cin>>N>>C; for(int i=1; i<=N; i++) { scanf("%d",&x[i]); } sort(x+1,x+1+N); int l=0,r=x[N],mid; while(l+1<r) { mid=(l+r)/2; if(fun(mid)) l=mid; else r=mid; } cout<<l<<endl; return 0; }<iostream> #include<algorithm> #include<cstdio> using namespace std; int x[100010]; int N,C; bool fun(int mid) { int cnt=1; int m=x[1]; for(int i=2; i<=N; i++) { if(x[i]-m>=mid) { cnt++; m=x[i]; } if(cnt>=C) return true; } return false; } int main() { cin>>N>>C; for(int i=1; i<=N; i++) { scanf("%d",&x[i]); } sort(x+1,x+1+N); int l=0,r=x[N],mid; while(l+1<r) { mid=(l+r)/2; if(fun(mid)) l=mid; else r=mid; } cout<<l<<endl; return 0; }