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
HintFor 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;
}