hdu 4932 BestCoder Round #4 1002

这题真是丧心病狂，引来今天的hack狂潮~

Miaomiao's Geometry

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 10    Accepted Submission(s): 3

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.

 

正解还在讨论中，不过有多种，，，被hack掉的。。

后来发现正解是暴力，暴力中的暴力。。。

枚举所有两点之间距离以及距离的一半，然后贪心的放。

对于A[i]，能往左放就往左，不行就往右。往右也放不了，就不行了，判断下一个长度。

 

#include <iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<string>
#include<cmath>
#include<queue>
#define N 1005
#define MAXN 1000005
#define M 200
#define mod 1000000007
#define ll long long
using namespace std;

int n,m;
int T;
ll a[N],b[N];
ll te;
ll ans;

bool cmp(ll x,ll y)
{
    return x>y;
}

int isok(ll v)
{
    ll now=0;
    for(int i=2;i<=n-1;i++){
       // printf(" %I64d %I64d %I64d now=%I64d v=%I64d
",a[i+1],a[i],a[i-1],now,v);
       // printf(" %d
",(a[i]-a[i-1]-now) <=v);
        if( (a[i]-a[i-1]-now) >=v){
            now=0;
        }
        else{
            if( (a[i+1]-a[i]) ==v){
                now=0;
                i++;
            }
            else if( (a[i+1]-a[i]) >v){
                now=v;
            }
            else return 0;
        }
    }
    return 1;
}

int main()
{
    int i,j,k;
    //freopen("data.txt","r",stdin);
    scanf("%d",&T);
    //while(scanf("%d",&n)!=EOF)
    while(T--)
   // for(cnt=1;cnt<=T;cnt++)
    {
       scanf("%d",&n);
       m=0;
       for(i=1;i<=n;i++){
            scanf("%I64d",&a[i]);
            a[i]*=4;
       }
       sort(a+1,a+1+n);
       //for(i=1;i<=n;i++)printf(" %I64d
",a[i]);
       for(i=1;i<n;i++){
            te=a[i+1]-a[i];
            b[m]=te;m++;
            te/=2;
            b[m]=te;m++;
       }
       sort(b,b+m,cmp);
       //for(i=0;i<m;i++)printf(" %I64d
",b[i]);
       for(j=0;j<m;j++){
            if(isok(b[j])==1){
                printf("%.3f
",(double)b[j]/4);
                break;
            }
       }

    }

    return 0;
}