HDU 4932 Miaomiao's Geometry（BestCoder Round #4）

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：有一些点，现在需要一些相同长度的片段遮住这些点（每个点在片段的两端才算遮住），这些片段不能有重叠部分，求出满足这样的片段的最大长度。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<queue>
#include<algorithm>
using namespace std;

const int N=1e5+10;
const int M=50000;
const int INF=0x3f3f3f3f;

double a[100], b[200]; ///b数组保存两点之间的长度及其一半
int n;

int Judge(double d) ///判断d能不能当满足条件的片段长度
{
    double p = a[0];
    int i;

    for (i = 1; i < n; i++)
    {
        if (fabs(p-a[i]) < 1e-9) continue;

        if (p > a[i]) return 0;
        else if (p+d <= a[i]) p = a[i];
        else p = a[i]+d;
    }

    return 1;
}

int main ()
{
    double Max;
    int T, i, k;

    scanf("%d", &T);

    while (T--)
    {
        scanf("%d", &n);
        for (i = 0; i < n; i++)
            scanf("%lf", &a[i]);

        memset(b, 0, sizeof(b));
        Max = -INF;
        k = 0;

        sort(a, a+n);

        for (i = 1; i < n; i++)
        {
            b[k++] = a[i]-a[i-1];
            b[k++] = (a[i]-a[i-1])/2;
        }

        for (i = 0; i < k; i++)
        {
            if (Judge(b[i]))
                Max = max(Max, b[i]);
        }

        printf("%.3f\n", Max);
    }

    return 0;
}

HDU 4932 Miaomiao's Geometry（BestCoder Round #4）