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

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.
 

 

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
",ans);
    }
    return 0;
}