首页 > 代码库 > Miaomiao's Geometry

Miaomiao's Geometry

2024-07-16 21:13:32   219人阅读

　　HDU 4932  Bestcoder

 

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.
 
 1 #include <stdio.h> 2 #include <string.h> 3 #include <algorithm> 4 #include<iostream> 5 #include<math.h> 6  7 using namespace std; 8  9 int main()10 {11     int cas,i,n,right,left;12     double res,a[55],b[120];13     cin>>cas;14     while(cas--)15     {16         cin>>n;17         int j=0;18         for(i=0;i<n;i++)19         {20             cin>>a[i];21         }22         sort(a,a+n);23         for(i=1;i<n;i++)24         {25             b[j++]=a[i]-a[i-1];26             b[j++]=(a[i]-a[i-1]) /2 ;27         }28         sort(b,b+j);29         int flag=0;30         j=j-1;31         res=(double)b[j];32 33         while(1)34         {35             right =0; left=0;36             flag=0;37             for(i=1;i<n;i++)38             {39                 if(i==n-1) continue;40                 if(a[i]-res<a[i-1] && a[i]+res>a[i+1])41                 {42                     flag=1;43                     break;44                 }45                 if(a[i]-res>=a[i-1]) 46                 {47                     if(right==1)48                     {49                         if(a[i]-a[i-1]==res) {left=1; right=0; }50                         else if(a[i]-a[i-1]>=2*res) { left=1; right=0; }51                         else if(a[i]+res<=a[i+1]) { left=0; right=1; }52                         else flag=1;53                     }54                     else { left=1; right=0; }55                 }56                 else if(a[i]+res<=a[i+1]) {57                     right=1;58                     left=0;59                 }60                 61             }62             if(flag==1) {63                 j--;64                 res=b[j];65             }66             else 67             {68                 printf("%.3lf\n",res);69                 break;70             }71         }72     }73     return 0;74 }

 

 
 
 
 
 
 
 
 

 


联系
我们