2020 cattle off winter training camp algorithm base 5 game to beef clan

https://ac.nowcoder.com/acm/contest/3006/B

The meaning of problems

　　Because beef clan often away matches, so a lot of the country to establish training bases, each base has a coordinate (the X-, the y-) .

　　This weekend, Taurus team have to go out of the game, each match point game in  x  -axis. Taurus team for the convenience of the game, looking for a training base to reach the maximum and minimum distances as a place to race.

Code

Bipartite

　　How to determine the current maximum distance is not a feasible solution, you can find the current maximum determined distance range for each point whether there is an intersection of the x (circle equation)

#include<bits/stdc++.h>
using namespace std;
const int maxn=100005;
struct node
{
    int x,y;
}point[maxn];
int n;
int check(double r);
int main()
{
    int i;
    double l=0,r=0x7fffffff,mid,ans;

    scanf("%d",&n);
    for(i=1;i<=n;i++)
        scanf("%d%d",&point[i].x,&point[i].y);
    
    while(r-l>1e-8)
    {
        mid=(l+r)/2;
        if(check(mid)) ans=mid,r=mid;
        else l=mid;
    }
    printf("%lf",ans);
    system("pause");
    return 0;
}
int check(double R)
{
    double l=-(0x7fffffff),r=0x7fffffff,gap;
    for(int i=1;i<=n;i++)
    {
        if(R<fabs(point[i].y))
            return 0;
        gap=sqrt(R*R-point[i].y*point[i].y);
        l=max(l,point[i].x-gap);
        r=min(r,point[i].x+gap);
    }
    return r>=l;
}

Thirds

　　template.

#include<bits/stdc++.h>
using namespace std;
const int maxn=100005;
struct node
{
    int x,y;
}point[maxn];
int n;
double check(double r);
int main()
{
    int i;
    double l=-(0x7fffffff),r=0x7fffffff,mid,mmid;

    scanf("%d",&n);
    for(i=1;i<=n;i++)
        scanf("%d%d",&point[i].x,&point[i].y);
    
    for(i=0;i<100;i++)
    {
      mid=(r+l)/2;
      mmid=(r+mid)/2;
      if(check(mid)>check(mmid))
        l=mid;
      else 
        r=mmid;
    }
    printf("%lf",check(mid));
    system("pause");
    return 0;
}
double check(double R)
{
    double maxn=0;
    for(int i=1;i<=n;i++)
      maxn=max(maxn,sqrt(point[i].y*point[i].y+(point[i].x-R)*(point[i].x-R)));
    return maxn;
}