三分·三分求极值

三分求极值

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

这一次我们就简单一点了，题目在此：

在直角坐标系中有一条抛物线y=ax^2+bx+c和一个点P(x,y)，求点P到抛物线的最短距离d。

输入

第1行：5个整数a,b,c,x,y。前三个数构成抛物线的参数，后两个数x,y表示P点坐标。-200≤a,b,c,x,y≤200

输出

第1行：1个实数d，保留3位小数(四舍五入)

样例输入

2 8 2 -2 6

样例输出

2.437

#include <iostream>
#include <math.h>
#include <float.h>
#include <iomanip>
using namespace std;
double presult = 0.0;
typedef struct funcparamter
{
    double a;
    double b;
    double c;
} funcparamter;

typedef struct point
{
    double x;
    double y;
} point;

double Calculator(funcparamter minefunc,point knownpoint,point foundpoint)
{
    return sqrt((foundpoint.x-knownpoint.x)*(foundpoint.x-knownpoint.x)+(minefunc.a*pow(foundpoint.x,2)+minefunc.b*foundpoint.x+minefunc.c-knownpoint.y)*(minefunc.a*pow(foundpoint.x,2)+minefunc.b*foundpoint.x+minefunc.c-knownpoint.y));
}
double findY(double x,funcparamter minefunc)
{
    return minefunc.a*pow(x,2)+minefunc.b*x+minefunc.c;
}

double findmin(double left,double right,funcparamter minefunc,point knownpoint)
{
    point lm,rm;
    lm.x = left+(right-left)/3;
    rm.x = left+2*(right-left)/3;
    lm.y = findY(lm.x,minefunc);
    rm.y = findY(rm.x,minefunc);

    if((Calculator(minefunc,knownpoint,lm)-Calculator(minefunc,knownpoint,rm))==0)
    {
        double min = Calculator(minefunc,knownpoint,lm)<Calculator(minefunc,knownpoint,rm) ?  Calculator(minefunc,knownpoint,lm) : Calculator(minefunc,knownpoint,rm);
        cout<<setiosflags(ios::fixed)<<setprecision(3)<<min<<endl;
        return min;
    }
    else if(Calculator(minefunc,knownpoint,lm)<Calculator(minefunc,knownpoint,rm))
    {
        findmin(left,rm.x,minefunc,knownpoint);
    }
    else
    {
        findmin(lm.x,right,minefunc,knownpoint);
    }
}
int main()
{
    funcparamter minefunc;
    point knownpoint;

    //cout<<DBL_MAX<<endl;
    cin>>minefunc.a>>minefunc.b>>minefunc.c>>knownpoint.x>>knownpoint.y;
    //cout<<Calculator(minefunc,knownpoint,foundpoint)<<endl;
    findmin(-1000,1000,minefunc,knownpoint);
    return 0;
}

提示