HDU 4932 贪心

Miaomiao's Geometry

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

Total Submission(s): 191    Accepted Submission(s): 38

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

 

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简直奇妙,比赛的时候900多就21个过的...,自己当时没考虑到一条线段能覆盖两个点的情况,说究竟还是自己太弱了,不够细心,还有就是自己太心急了,刚敲完就交了,导致罚时比較多,今后得慢慢改,注意到答案仅仅能是距离或者距离的一半,依次枚举即可,对每一个点仅仅有两种选择,一种是选点左边的线段,一种是选右边的线段,当能选左边的时候一定要选左边的,否则选右边的,如果左右两边都不能选,那么这个线段肯定长了,如果当前枚举的距离为x,那么选左边的条件是A[j]-A[j-1]-vis[j]>=x||A[j]==A[j-1]+x,右边这样的就是一条线段覆盖两个点的情况,vis[j]是上一个点对如今这个区间的影响.

代码例如以下:

#include <iostream>
#include <map>
#include<algorithm>
#include <stack>
#include <string.h>
#include <queue>
#include<cstdio>
using namespace std;
#define INF  5e9+7
typedef long long LL;
int main()
{
    //freopen("in.txt","r",stdin);
    long long T,N;
    double A[100];
    double vis[100];
    cin>>T;
    while(T--)
    {
        cin>>N;
        for(int i=1; i<=N; i++)cin>>A[i];
        sort(A+1,A+N+1);
        double ans=0;
        A[N+1]=INF;
        for(int i=1; i<=N-1; i++)
        {
            memset(vis,0,sizeof(vis));
            double x=A[i+1]-A[i];
            bool ok1=true,ok2=true;
            for(int j=2; j<=N-1; j++)
            {
                if((A[j]-A[j-1]-vis[j]>=x)||(A[j]==A[j-1]+x))
                {
                    continue;
                }
                if(A[j+1]-A[j]>=x)
                {
                    vis[j+1]=x;
                    continue;
                }
                ok1=false;
                break;
            }
            if(ok1)
            {
                ans=max(x,ans);
            }
            memset(vis,0,sizeof(vis));
            for(int j=2; j<=N-1; j++)
            {
                if(A[j]-A[j-1]-vis[j]>=x/2||A[j]==A[j-1]+x/2)
                {
                    continue;
                }
                if(A[j+1]-A[j]>=x/2)
                {
                    vis[j+1]=x/2;
                    continue;
                }
                ok2=false;
                break;
            }
            if(ok2)ans=max(ans,x/2);
        }
        printf("%.3f\n",ans);
    }
    return 0;
}