[POJ2456]Aggressive cows

 

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8760		Accepted: 4350

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

THINKING

　　最大化最小值的题。经分析，我们可以设C(D)=可以安排牛的位置使得任意的牛的间距都不小d，则用贪心法可以求解。首先对牛舍位置排序，把第一头牛放入x0的牛舍，如果第i头牛放入牛舍aj的话，第i+1头牛就要放入aj<=ak中，二分查找答案。

{《挑战ACM程序设计竞赛》}
var n,m,lb,ub,mid:int64;
    i:longint;
    a:array[-1..10000] of int64;

function c(x:int64):boolean;
var last,crt:int64;j:longint;
begin
    last:=0;
    //crt:=0;
    for j:=1 to m-1 do
        begin
            crt:=last+1;
            while (crt<n)and(a[crt]-a[last]<x) do inc(crt);
            if crt=n then exit(false);
            last:=crt;
        end;
    exit(true);
end;

procedure sort(l,r: int64);
      var
         i,j:longint;x,y:int64;
      begin
         i:=l;
         j:=r;
         x:=a[(l+r) div 2];
         repeat
           while a[i]<x do
            inc(i);
           while x<a[j] do
            dec(j);
           if not(i>j) then
             begin
                y:=a[i];
                a[i]:=a[j];
                a[j]:=y;
                inc(i);
                j:=j-1;
             end;
         until i>j;
         if l<j then
           sort(l,j);
         if i<r then
           sort(i,r);
      end;

begin
    readln(n,m);
    for i:=1 to n do readln(a[i]);
    sort(1,n);
    lb:=0;
    ub:=a[n]-a[1]+1;
    while ub-lb>1 do
        begin
            mid:=(lb+ub) div 2;
            if c(mid) then lb:=mid
            else ub:=mid;
        end;
    writeln(lb-1);
end.