HDU 4932 Miaomiao's Geometry 暴力

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Miaomiao's Geometry

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 438    Accepted Submission(s): 107

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

331 2 331 2 441 9 100 10

 

Sample Output

1.0002.0008.000
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

 

有n个点在x坐标上，要求用一些线段将这些点全部覆盖，要求这些线段不能重叠，且必须等长，x坐标上的点必须在线段的端点，求线段的最长长度。

一开始从第二个点到倒数第二个点找左右距离最大的值存到数组里面，最后输出此数组最小的即可，也就是求个最大最小。但是如果数据是这样的：2 7 8 14 15的话，果断WA，因为8和14共用一段区间，且此区间要大于2-7的5,。

正确解法应该是从大到小枚举将这些数的差值以及差值的一半，如果符合要求，则就是答案。

//15MS268K#include<stdio.h>#include<algorithm>using namespace std;double s[57],array[107];int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,k=0;        scanf("%d",&n);        for(int i=0;i<n;i++)            scanf("%lf",&s[i]);        sort(s,s+n);        for(int i=1;i<n;i++)        {            array[k++]=s[i]-s[i-1];            array[k++]=(s[i]-s[i-1])/2;        }        sort(array,array+k);        for(int i=k-1;i>=0;i--)        {            bool flag=1;            double last=s[0],a=array[i];            for(int j=1;j<n-1;j++)            {                if(last==s[j])continue;                if(last+a<=s[j])last=s[j];                else if(s[j]+a<=s[j+1])last=s[j]+a;                else {flag=0;break;}            }            if(flag){printf("%.3lf\n",a);break;}        }    }    return 0;}