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

 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=;
        for(i=;i<=n;i++){
             scanf("%I64d",&a[i]);
             a[i]*=;
        }
        sort(a+,a++n);
        //for(i=1;i<=n;i++)printf(" %I64d\n",a[i]);
        for(i=;i<n;i++){
             te=a[i+]-a[i];
             b[m]=te;m++;
             te/=;
             b[m]=te;m++;
        }
        sort(b,b+m,cmp);
        //for(i=0;i<m;i++)printf(" %I64d\n",b[i]);
        for(j=;j<m;j++){
             if(isok(b[j])==){
                 printf("%.3f\n",(double)b[j]/);
                 break;
             }
        }

     }

     return ;
 }