POJ 2456 Aggressive cows

Aggressive cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9797		Accepted: 4865

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.

最大化最小值

农夫约翰搭了一间有n间牛舍的小屋。牛舍排在一条线上第i号牛舍在xi的位置。但是他的m头牛对小屋很不满意，因此经常互相攻击。约翰为了防止牛之间互相伤害，因此决定把每头牛都放在离其他牛尽可能远的牛舍。也就是要最大化最近的两头牛之间的距离

分析：

我们定义C(d):=可以安排牛的位置是的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)的最大值的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) = 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非常容易的求解。

1.       对牛舍的位置x进行排序

2.       把第一头牛放入x0的牛舍

3.       如果第i头牛放入了xj的话，第i+ 1头牛就要放入满足xj + d <= xk的最小的xk中

对x的排序只需在最开始时进行一次就可以了，每一次判断对每头牛最多进行一次处理，因此复杂度是O(N)。

//最大化最小值
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 100000 + 10;
const int INF = 100000000;

int n, m;
int x[maxn];

bool C(int d)
{
    int last = 0;
    for (int i = 1; i < m; i++){
        int crt = last + 1;
        while (crt < n && x[crt] - x[last] < d){
            crt++;
        }
        if  (crt == n)
            return false;
        last = crt;
    }
    return true;
}

void solve()
{
    //从最开始对x数组排序
    sort(x, x + n);

    //初始化解的存在范围
    int lb = 0, ub = INF;

    while (ub - lb > 1){
        int mid = (lb + ub) / 2;
        if (C(mid))
            lb = mid;
        else
            ub = mid;
    }
    printf("%d\n", lb);
}

int main()
{
    while (scanf("%d%d", &n, &m) != EOF){
        for (int i = 0; i < n; i++){
            scanf("%d", &x[i]);
        }
        solve();
    }
    return 0;
}