程序设计与算法（二）算法基础》《第四周二分》Aggressive cows 2456

小咪咪 2023-08-17 16:23 122阅读 0赞

## 2456:Aggressive cows ##

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 *  [提交][Link 2]
 *  [统计][Link 3]
 *  [提示][Link 4]
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总时间限制:

1000ms

内存限制:

65536kB

描述

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?

输入

\* Line 1: Two space-separated integers: N and C  
  
\* Lines 2..N+1: Line i+1 contains an integer stall location, xi

输出

\* Line 1: One integer: the largest minimum distance

样例输入

5 3
    1
    2
    8
    4
    9

样例输出

提示

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.

题意要表达的是：把C头牛放到N个带有编号的隔间里，使得任意两头牛所在的隔间编号的最小差值最大。

分析：这是一个最小值最大化的问题。先对隔间编号从小到大排序，则最大距离不会超过两端的两头牛之间的差值，最小值为0。所以我们可以通过二分枚举最小值来求。假设当前的最小值为x，如果判断出最小差值为x时可以放下C头牛，说明当前的x有点小，就先让x变大再判断；如果放不下，说明当前的x太大了，就先让x变小然后再进行判断。直到求出一个最大的x就是最终的答案。

#define _CRT_SECURE_NO_WARNINGS
    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    
    using namespace std;
    
    int a[100006];
    int N, C;
    bool func(int x)
    {
        int temp = a[0];
        int cnt = 1;
        for (int i = 1; i < N; i++)
        {
            //统计在设定间距下（比如4），牛舍的位置是否放的下C头牛
            if (a[i] - temp >= x)
            {
                cnt++;
                temp = a[i];
            }
            if (cnt == C)
            {
                return true;
            }
        }
        return false;
    }
    
    int main()
    {
        int i;
        cin >> N; // room number for cows
        cin >> C; // cows number
        for (i = 0; i < N; i++)
        {
            scanf("%d", &a[i]);
        }
        sort(a, a + N);
    
        int  left, right, mid;
        left = 0;
        right = a[N - 1]-a[0];
        //mid = left + (right - left) / 2;
        while (left <= right)
        {
            // 5个牛舍的位置【1， 2， 4， 8， 9】，最大间距是 8，所有可能间距取值是【0~8】，二分查找问题
            mid = left + (right - left) / 2;//第一把间距是4，失败，需要缩小间距
            if ((func(mid)))  // 如果现在的间距满足放下C头牛，那要试一试间距变大的情况
            { // 间距如何变大？
                left = mid + 1; /* left = 5 ,[0~4]delete,  [5~8]select*/
            }
            else
            {
            // 间距缩小
                right = mid - 1; /*right = 3, [3~8]delete,  [0~3]select */
            }
        }
        // 当间距=3时，满足成立条件，return true, 此时，left=3+1=4, while(4<=3)退出，实际间距4-1=3
        printf("%d\n",left-1);
    
        return 0;
    }

转载于:https://www.cnblogs.com/focus-z/p/11521900.html

[Link 1]: http://bailian.openjudge.cn/practice/2456/
[Link 2]: http://bailian.openjudge.cn/practice/2456/submit/
[Link 3]: http://bailian.openjudge.cn/practice/2456/statistics/
[Link 4]: http://bailian.openjudge.cn/practice/hint/2456/
[Link 5]: http://bailian.openjudge.cn/practice/clarify/2456/