hdu 4932 Miaomiao's Geometry（暴力枚举）

pid=4932">Miaomiao's Geometry

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

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

3
3
1 2 3
3
1 2 4
4
1 9 100 10

 

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.

 

题意：给出n个点，找出一些等长的线段覆盖这些点。这些点仅仅能作为线段的端点，并且随意两条线段的相交长度不能大于0.求满足条件的线段的最大长度。

分析：通过分析能够得出。终于结果是相邻两点之间的长度，或者相邻两点之间长度的一半。由于最多仅仅有50个点，100个长度，所以仅仅需枚举这些长度。求出一个满足条件的最长线段就可以。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
    double b[120], c[60];
    int flag[60];
    int n, i, j, T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(i = 0; i < n; i++)
            scanf("%lf",&c[i]);
        sort(c, c+n);
        int m = 0;
        for(i = 1; i < n; i++)
        {
            b[m++] = c[i] - c[i-1];
            b[m++] = (c[i] - c[i-1]) / 2;
        }
        sort(b, b+m);
        double ans;
        for(i = m - 1; i >= 0; i--)
        {
            memset(flag, 0, sizeof(flag));
            flag[0] = 1;
            double tmp = b[i];
            for(j = 1; j < n - 1; j++)
            {
                if(c[j] - tmp < c[j-1] && c[j] + tmp > c[j+1])  //往左往右都不行
                    break;
                if(c[j] - tmp >= c[j-1])
                {
                    if(flag[j-1] == 2) // 前一个往右
                    {
                        if(c[j] - c[j-1] == tmp) flag[j] = 1;  //两个点作为线段的两个端点
                        else if(c[j] - c[j-1] >= 2 * tmp) flag[j] = 1; //一个往左。一个往右
                        else if(c[j] + tmp <= c[j+1]) flag[j] = 2; //仅仅能往右
                        else break;
                    }
                    else flag[j] = 1;
                }
                else if(c[j] + tmp <= c[j+1])
                    flag[j] = 2;
            }
            if(j == n - 1)
            {
                ans = tmp;
                break;
            }
        }
        printf("%.3lf\n", double(ans));
    }
    return 0;
}