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

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4932

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

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

 

题意：

求最大能够覆盖所有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，然后依次从最大的开始枚举进行验证即可。

代码如下：

#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;//无论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)//如果前一个是向右的            {                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;//只能向右                }                else                    return false;            }            else                f[i] = 1;        }        else if(p[i] + tt <= p[i+1])            f[i] = 2;    }    if(i == n-1)//全部符合        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; //开始肯定是让线段向左            if(Judge(d[i]))            {                ans = d[i];                break;            }        }        printf("%.3lf\n",ans);    }    return 0;}