POJ 2456 Aggressive cows （二分搜索）

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

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.

分析：
类似最大化最小值或者最小化最大值的问题，通常用二分搜索法就可以很好的解决。
定义：C（d）：表示，可以安排牛的位置使得最近的两头牛的距离不小于d

那么问题就变成了求满足C(d)的最大的d。另外，最近的间距不小于d也可以说成是所有牛的间距都不小于d，因此就有，
C(d)，表示，可以安排牛的位置使得任意的牛的间距都不小于d
这个问题的判断使用贪心法便可以很好的解决。

1.对牛舍的位置x进行排序
2.把第一头牛放入x0的位置
3.如果第i头牛放入了xj的话，第i+1头牛要放入满足xj+d的最小的xk中，

对x的排序只需要在最开始的时候进行一次就可以了，每一次判断对每头牛最多进行一次处理。

代码：

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
int N,M;
int x[100009];
bool C(int d)
{
    int last=0;
    for(int i=1; i<M; i++)///看看这个距离能不能放下
    {
        int crt=last+1;
        while(crt<N&&x[crt]-x[last]<d)
            crt++;
        if(crt==N) return false;
        last=crt;
    }
    return true;
}

void solve()
{
    sort(x,x+N);
    int lb=0,ub=0x3f3f3f3f;
    while(ub-lb>1)
    {
        int mid=(lb+ub)/2;
        if(C(mid)) lb=mid;
        else ub=mid;
    }
    printf("%d
",lb);
}

int main()
{
    scanf("%d%d",&N,&M);
    for(int i=0; i<N; i++)
        scanf("%d",&x[i]);
    solve();
    return 0;
}

POJ 2456 Aggressive cows （ 二分搜索）

POJ 2456 Aggressive cows （二分搜索）