CF-51C - Three Base Stations（二分或枚举优化）

C - Three Base Stations

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

The New Vasjuki village is stretched along the motorway and that's why every house on it is characterized by its shift relative to some fixed point — the x_i coordinate. The village consists of n houses, the i-th house is located in the point with coordinates of x_i.

TELE3, a cellular communication provider planned to locate three base stations so as to provide every house in the village with cellular communication. The base station having power d located in the point t provides with communication all the houses on the segment[t - d, t + d] (including boundaries).

To simplify the integration (and simply not to mix anything up) all the three stations are planned to possess the equal power of d. Which minimal value of d is enough to provide all the houses in the village with cellular communication.

Input

The first line contains an integer n (1 ≤ n ≤ 2· 10⁵) which represents the number of houses in the village. The second line contains the coordinates of houses — the sequence x₁, x₂, ..., x_n of integer numbers (1 ≤ x_i ≤ 10⁹). It is possible that two or more houses are located on one point. The coordinates are given in a arbitrary order.

Output

Print the required minimal power d. In the second line print three numbers — the possible coordinates of the base stations' location. Print the coordinates with 6 digits after the decimal point. The positions of the stations can be any from 0 to 2· 10⁹ inclusively. It is accepted for the base stations to have matching coordinates. If there are many solutions, print any of them.

Sample Input

Input

4
1 2 3 4

Output

0.500000
1.500000 2.500000 3.500000

Input

3
10 20 30

Output

0
10.000000 20.000000 30.000000

Input

5
10003 10004 10001 10002 1

Output

0.500000
1.000000 10001.500000 10003.500000

本题有两种思路：一种，枚举三段，找最小，然后优化，第二种，使用二分距离判断是否符合条件。本题有个隐藏条件，那就最终的距离只可能是前面一个整数，后面跟个0.5或0.

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int mm=2e5+9;
const int oo=1e9;
int dis[mm];int n;
double a,b,c,len=oo;
void ok(int x,int y)
{ if(y<0)y=0;
  int z;
  z=dis[x]-dis[0];
  z=max(dis[y]-dis[min(x+1,n-1)],z);
  z=max(dis[n-1]-dis[min(y+1,n-1)],z);
  if(z<len)
  {
    len=z;a=dis[x]+dis[0];b=dis[y]+dis[min(x+1,n-1)];c=dis[n-1]+dis[min(y+1,n-1)];
  }
}
int main()
{

    scanf("%d",&n);
   len=oo;
    for(int i=0;i<n;i++)
      scanf("%d",&dis[i]);
    sort(dis,dis+n);
    for(int i=0,j=0;i<n;i++)
    { 
      while(j<n-1&&dis[n-1]-dis[j+1]>dis[j]-dis[i+1])
        ++j;
      ok(i,j);ok(i,j-1);
    }
    len/=2.0;a/=2.0;b/=2.0;c/=2.0;
   printf("%.6lf\n%.6lf %.6lf %.6lf\n",len,a,b,c);

}

二分代码

#include<iostream>
#include<cstring>
#include<iomanip>
#include<algorithm>
using namespace std;
const int mm=2e5+9;
const double oo=1e39;
const double ex=1e-7;
double locat[mm];
int n,ans[mm];
int blook(int x)///返回比x大的第一个数的下标
{
  int l=0,r=n,mid=(l+r)/2;
  while(l<r)
  {
    if(locat[mid]>x)r=mid;
    else l=mid+1;
    mid=(l+r)/2;
  }
  return mid;
}
bool ok(int x)
{ int z=0;
  for(int i=0;i<3;i++)
  {
    z=blook(locat[z]+x);
    ans[i]=z;
    if(z==n)return 1;
  }
  return 0;
}

int main()
{
  while(cin>>n)
  {
    for(int i=0;i<n;i++)
      cin>>locat[i];
    sort(locat,locat+n);
    int r=2e9+9,l=0,mid;
    while(l<r)
    {
      mid=(l+r)/2;
      if(ok(mid))r=mid;
      else l=mid+1;
    }
    ok(l);
    double len=l;len/=2.0;
    double a,b,c;a=locat[ans[0]-1]+locat[0];b=locat[ans[1]-1]+locat[ans[0]];
    c=locat[ans[2]-1]+locat[ans[1]];
    cout.setf(ios::fixed);
    cout<<setprecision(6)<<len<<"\n"<<a/2.0<<" "<<b/2.0<<" "<<c/2.0<<endl;
  }
}