HDU4932 Miaomiao's Geometry

Miaomiao's Geometry

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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.

题意：在x轴上有一些不重复的点，要求用线段覆盖它们，覆盖的规则是：

1、线段的左断点或右端点落在点上才算覆盖这个点

2、线段不能重叠

问最长的线段是多长？

题解：这个长度肯定是两个点之间的距离，或者距离的一半。所以枚举这个距离，判断是否符合要求就行了，记录下最大的值。判断时用贪心，先判断点的左边能不能放线段，不能的话就放在右边，都不能的话，则不满足。最两端的端点可以不用判断。

代码：

#include <stdio.h>#include <algorithm>using namespace std;#define INF 2e9+5double a[55], b[105];bool judge(double x, int n){int flag = 0;for(int i = 1; i < n-1; i++){if(flag == 0){if(a[i] - a[i-1] < x){flag = 1;if(a[i+1]-a[i] < x)return 0;}}else{if(a[i] - a[i-1] >= x*2 || a[i] - a[i-1] == x)flag = 0;else{if(a[i+1] - a[i] < x)return 0;}}}return 1;}int main(){int t, n;scanf("%d", &t);while(t--){scanf("%d", &n);for(int i = 0; i < n; i ++){scanf("%lf", &a[i]);}sort(a, a+n);for(int i = 1; i < n; i++){b[i] = a[i] - a[i-1];}sort(b+1, b+n);double ans = 0;for(int i = 1; i < n; i++){if(i > 0 && b[i] == b[i-1])continue;if(judge(b[i], n))ans = max(ans, b[i]);else if(judge(b[i]/2, n))ans = max(ans, b[i]/2);}printf("%.3lf\n", ans);}return 0;}