Miaomiao's Geometry

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 410 Accepted Submission(s): 147

Problem Description

There are N point on X-axis . Miaomiao would like to cover them ALL by using segments with same length.

There are 2 limits:

1.A point is convered if there is a segments T , the point is the left end or the right end of T.

2.The length of the intersection of any two segments equals zero.

For example , point 2 is convered by [2 , 4] and not convered by [1 , 3]. [1 , 2] and [2 , 3] are legal segments , [1 , 2] and [3 , 4] are legal segments , but [1 , 3] and [2 , 4] are not (the length of intersection doesn't equals zero), [1 , 3] and [3 , 4]
are not(not the same length).

Miaomiao wants to maximum the length of segements , please tell her the maximum length of segments.

For your information , the point can't coincidently at the same position.

Input

There are several test cases.

There is a number T ( T <= 50 ) on the first line which shows the number of test cases.

For each test cases , there is a number N ( 3 <= N <= 50 ) on the first line.

On the second line , there are N integers Ai (-1e9 <= Ai <= 1e9) shows the position of each point.

Output

For each test cases , output a real number shows the answser. Please output three digit after the decimal point.

Sample Input

Sample Output

1.000

2.000

8.000

Hint

For the first sample , a legal answer is [1,2] [2,3] so the length is 1.

For the second sample , a legal answer is [-1,1] [2,4] so the answer is 2.

For the thired sample , a legal answer is [-7,1] , [1,9] , [10,18] , [100,108] so the answer is 8.

Source

BestCoder Round #4

题意：

求最大可以覆盖全部所给的点的区间长度（所给的点必须处于区间两端）。

思路：

答案一定是相邻点之间的差值或者是相邻点之间的差值除以2，那么把这些可能的答案先算出来。然后依次从最大的開始枚举进行验证就可以。

代码例如以下：

#include <cstdio>

#include <algorithm>

#include <cstring>

using namespace std;

const int MAXN = 147;

int f[MAXN];//记录线段方向

double p[MAXN];

double d[MAXN];//相邻断点的差值

int n;

void init()

{

    memset(p,0,sizeof(p));

    memset(f,0,sizeof(f));

    memset(d,0,sizeof(d));

}

bool Judge(double tt)

{

     int i;

    for(i = 1; i < n-1; i++)

    {

        if(p[i] - tt < p[i-1] && p[i] + tt > p[i+1])

            break;//不管向左还是向右均为不符合

        if(p[i] - tt >= p[i-1])//向左察看

        {

            if(f[i-1] == 2)//假设前一个是向右的

            {

                if(p[i] - p[i-1] == tt)

                    f[i] = 1;//两个点作为线段的两个端点

                else if(p[i] - p[i-1] >= 2*tt)//一个向左一个向右

                {

                    f[i] = 1;

                }

                else if(p[i] + tt <= p[i+1])

                {

                    f[i] = 2;//仅仅能向右

                }

                else

                    return false;

            }

            else

                f[i] = 1;

        }

        else if(p[i] + tt <= p[i+1])

            f[i] = 2;

    }

    if(i == n-1)//所有符合

        return true;

    return false;

}

int main()

{

    int t;

    scanf("%d",&t);

    while(t--)

    {

        init();

        scanf("%d",&n);

        for(int i = 0; i < n; i++)

        {

            scanf("%lf",&p[i]);

        }

        sort(p,p+n);

        int cont = 0;

        for(int i = 1; i < n; i++)

        {

            d[cont++] = p[i] - p[i-1];

            d[cont++] = (p[i] - p[i-1])/2.0;

        }

        sort(d,d+cont);

        double ans = 0;

        for(int i = cont-1; i >= 0; i--)

        {

            memset(f,0,sizeof(f));

            f[0] = 1; //開始肯定是让线段向左

            if(Judge(d[i]))

            {

                ans = d[i];

                break;

            }

        }

        printf("%.3lf\n",ans);

    }

    return 0;

}

hdu4932 Miaomiao's Geometry （BestCoder Round #4 枚举)

Miaomiao's Geometry