HDU 4932 Miaomiao's Geometry

Miaomiao's Geometry

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 379    Accepted Submission(s): 93

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.
 

 

Source

BestCoder Round #4

题目意思：

给你n个位置，你要用等长度的区间线段去覆盖他们，而且只能用端点去覆盖，不能出现位置点在区间线段内的情况。问最大长度是多少，位置点的范围是-1e9~1e9。

分析：由于线段条数只有50条，那么对于排序后的位置点从间距最大的长度开始暴力检查，但是注意，取一半的情况1 2 5 6 8这样的。

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<vector>#include<queue>#include<stack>#define rt return#define sf scanf#define pf printf#define si(n) sf("%d",&n)#define pi(n) pf("%d",n)#define REP0(i,n) for(int i=0;i<(n);i++)#define REP1(i,n) for(int i=1;i<=(n);i++)#define REP(i,s,n) for(int i=s;i<=(n);i++)#define db double#define pb push_back#define LL long long#define INF 0x3fffffff#define eps 1e-8#define PI acos(-1)#define maxn 55using namespace std;int n;double pos[maxn],pp[maxn];double len[maxn*2];int main(){    #ifdef ACBang    freopen("in.txt","r",stdin);    #endif    int T;sf("%d",&T);    while(T--){        sf("%d",&n);        REP0(i,n)            sf("%lf",&pos[i]);        sort(pos,pos+n);        for(int i=0;i<n;i++)                pp[i]=pos[i];        int k=0;        REP0(i,n-1){            len[k++]=abs(pos[i+1]-pos[i]);            len[k++]=len[k-1]*0.5;        }//        REP0(i,k)cout<<len[i]<<" ";cout<<endl;        sort(len,len+k);        int l=unique(len,len+k)-len;        for(int i=l-1;i>=0;i--){            for(int t=0;t<n;t++)                pos[t]=pp[t];            bool tmp=true;            for(int j=1;j<n-1;j++){                if(pos[j-1]<=pos[j]-len[i])                    continue;                if(pos[j]+len[i]<pos[j+1]){                    pos[j]=pos[j]+len[i];                    continue;                }                if(pos[j]+len[i]==pos[j+1]){                    continue;                }                tmp=false;                break;            }            if(tmp==true){                pf("%.3lf\n",len[i]);                break;            }        }    }    rt 0;}