BNU 4260 Trick or Treat && ZOJ 3386 (三分查找)

【题目链接】click here~~

【题目大意】求x轴上一点到各点的最大值中的最小值 点到线段距离 

【解题思路】三分查找

三分查找初学可以参考这篇博客分析：三分查找，写的很详细了，其实跟类似于二分查找，理解了如何构造，代码不难实现

方法1：

#include <bits/stdc++.h>
using namespace std;
const double eps=1e-7;
const double inf=0x3f3f3f3f;
const int N=55000;
int n;
struct point
{
    double x,y;
}mapp[N];
double dis(point a,point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double getmax(double x) //最长边，最长时间
{
    double maxx=eps;
    point q={x,0};
    for(int i=0;i<n;i++)
    {
        double p=dis(q,mapp[i]);
        maxx=max(maxx,p);
    }
    return maxx;
}
int main()
{
    //freopen("1.txt","r",stdin);
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0) break;
        double pleft=N,pright=-N;
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&mapp[i].x,&mapp[i].y);
            pleft=min(pleft,mapp[i].x);  //左边界
            pright=max(pright,mapp[i].x);//右边界
        }
        while((pright-pleft)>eps)
        {
            double mid=(pright-pleft)/3;
            double l,r;
            l=pleft+mid,r=pleft+2*mid;
            if(getmax(l)>=getmax(r)) pleft=l;
            else pright=r;
        }
        printf("%.9f %.9f\n",pleft,getmax(pleft));
    }
    return 0;
}

方法2：

#include <bits/stdc++.h>
using namespace std;
const int N=55000;
const double eps=1e-7;
int n,m;
struct point{
    double x,y;
}mapp[N];
double disget(point a,point b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double getmax(double  x){//求最大边，最长时间
    point q={x,0};
    double maxx=eps;
    for(int i=0;i<n;i++){
        double p=disget(q,mapp[i]);
        maxx=max(maxx,p);
    }
    return maxx;
}
double solve(){//赋初值时，最好直接赋数字！
    double pleft=-200000,pright=200000;
    double mid,midd,mid_area,midd_area ;
    while(pright-pleft>eps){
        mid=(pleft+pright)/2.0 ;
        midd=(mid+pright)/2.0 ;
        mid_area=getmax(mid) ;
        midd_area=getmax(midd) ;
        if(mid_area<=midd_area) pright=midd ;
        else pleft=mid ;
    }
    return midd ;
}
int main()
{
    while(scanf("%d",&n)!=EOF){
        if(n==0) break;
        for(int i=0;i<n;i++){
            scanf("%lf%lf",&mapp[i].x,&mapp[i].y);
        }
        double res=solve();
        printf("%.9f %.9f\n",res,getmax(res));
    }
    return 0;
}

BNU 4260 Trick or Treat && ZOJ 3386 (三分查找)

原文：http://blog.csdn.net/u013050857/article/details/45340393