三点确定一个圆(Apple HDU - 6206)(计算几何+高精度)

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三点确定一个圆的计算方法

设一个圆的圆心坐标为(x0,y),半径为r。那么这个圆的方程可以写成

(x-x0)^2+(y-y0)^2=r^2

在这个圆上随便取三个点,设这三个点的坐标分别为(x1,y1),(x2,y2),(x3,y3)。那么有

公式(1)(2)相减,(1)(3)相减经过化简可以得到:

x0,y0有唯一解的条件是系数行列式不为0:

简单变形就是:

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这样也就是三点不能共线

设:

那么结果

这样答案就出来了

题解 :先判断出圆心的位置,然后进而算出半径的大小,最后跟据最后一个点与这个圆心的距离进行判断大小,还使用了个大大大的大整数类,不会JAVA,赶快学


#include<bits/stdc++.h>

#define MAX_L 2005 //最大长度,可以修改

using namespace std;

#define max_n 100000000//因为结果可能不是整数,把小数部分去掉

class bign
{
public:
    int len, s[MAX_L];//数的长度,记录数组
//构造函数
    bign();
    bign(const char*);
    bign(int);
    bool sign;//符号 1正数 0负数
    string toStr() const;//转化为字符串,主要是便于输出
    friend istream& operator>>(istream &,bign &);//重载输入流
    friend ostream& operator<<(ostream &,bign &);//重载输出流
//重载复制
    bign operator=(const char*);
    bign operator=(int);
    bign operator=(const string);
//重载各种比较
    bool operator>(const bign &) const;
    bool operator>=(const bign &) const;
    bool operator<(const bign &) const;
    bool operator<=(const bign &) const;
    bool operator==(const bign &) const;
    bool operator!=(const bign &) const;
//重载四则运算
    bign operator+(const bign &) const;
    bign operator++();
    bign operator++(int);
    bign operator+=(const bign&);
    bign operator-(const bign &) const;
    bign operator--();
    bign operator--(int);
    bign operator-=(const bign&);
    bign operator*(const bign &)const;
    bign operator*(const int num)const;
    bign operator*=(const bign&);
    bign operator/(const bign&)const;
    bign operator/=(const bign&);
//四则运算的衍生运算
    bign operator%(const bign&)const;//取模(余数)
    bign factorial()const;//阶乘
    bign Sqrt()const;//整数开根(向下取整)
    bign pow(const bign&)const;//次方
//一些乱乱的函数
    void clean();
    ~bign();
};

#define max(a,b) a>b ? a : b
#define min(a,b) a<b ? a : b

bign::bign()
{
    memset(s, 0, sizeof(s));
    len = 1;
    sign = 1;
}

bign::bign(const char *num)
{
    *this = num;
}

bign::bign(int num)
{
    *this = num;
}

string bign::toStr() const
{
    string res;
    res = "";
    for (int i = 0; i < len; i++)
        res = (char)(s[i] + '0') + res;
    if (res == "")
        res = "0";
    if (!sign&&res != "0")
        res = "-" + res;
    return res;
}

istream &operator>>(istream &in, bign &num)
{
    string str;
    in>>str;
    num=str;
    return in;
}

ostream &operator<<(ostream &out, bign &num)
{
    out<<num.toStr();
    return out;
}

bign bign::operator=(const char *num)
{
    memset(s, 0, sizeof(s));
    char a[MAX_L] = "";
    if (num[0] != '-')
        strcpy(a, num);
    else
        for (int i = 1; i < strlen(num); i++)
            a[i - 1] = num[i];
    sign = !(num[0] == '-');
    len = strlen(a);
    for (int i = 0; i < strlen(a); i++)
        s[i] = a[len - i - 1] - 48;
    return *this;
}

bign bign::operator=(int num)
{
    char temp[MAX_L];
    sprintf(temp, "%d", num);
    *this = temp;
    return *this;
}

bign bign::operator=(const string num)
{
    const char *tmp;
    tmp = num.c_str();
    *this = tmp;
    return *this;
}

bool bign::operator<(const bign &num) const
{
    if (sign^num.sign)
        return num.sign;
    if (len != num.len)
        return len < num.len;
    for (int i = len - 1; i >= 0; i--)
        if (s[i] != num.s[i])
            return sign ? (s[i] < num.s[i]) : (!(s[i] < num.s[i]));
    return !sign;
}

bool bign::operator>(const bign&num)const
{
    return num < *this;
}

bool bign::operator<=(const bign&num)const
{
    return !(*this>num);
}

bool bign::operator>=(const bign&num)const
{
    return !(*this<num);
}

bool bign::operator!=(const bign&num)const
{
    return *this > num || *this < num;
}

bool bign::operator==(const bign&num)const
{
    return !(num != *this);
}

bign bign::operator+(const bign &num) const
{
    if (sign^num.sign)
    {
        bign tmp = sign ? num : *this;
        tmp.sign = 1;
        return sign ? *this - tmp : num - tmp;
    }
    bign result;
    result.len = 0;
    int temp = 0;
    for (int i = 0; temp || i < (max(len, num.len)); i++)
    {
        int t = s[i] + num.s[i] + temp;
        result.s[result.len++] = t % 10;
        temp = t / 10;
    }
    result.sign = sign;
    return result;
}

bign bign::operator++()
{
    *this = *this + 1;
    return *this;
}

bign bign::operator++(int)
{
    bign old = *this;
    ++(*this);
    return old;
}

bign bign::operator+=(const bign &num)
{
    *this = *this + num;
    return *this;
}

bign bign::operator-(const bign &num) const
{
    bign b=num,a=*this;
    if (!num.sign && !sign)
    {
        b.sign=1;
        a.sign=1;
        return b-a;
    }
    if (!b.sign)
    {
        b.sign=1;
        return a+b;
    }
    if (!a.sign)
    {
        a.sign=1;
        b=bign(0)-(a+b);
        return b;
    }
    if (a<b)
    {
        bign c=(b-a);
        c.sign=false;
        return c;
    }
    bign result;
    result.len = 0;
    for (int i = 0, g = 0; i < a.len; i++)
    {
        int x = a.s[i] - g;
        if (i < b.len) x -= b.s[i];
        if (x >= 0) g = 0;
        else
        {
            g = 1;
            x += 10;
        }
        result.s[result.len++] = x;
    }
    result.clean();
    return result;
}

bign bign::operator * (const bign &num)const
{
    bign result;
    result.len = len + num.len;

    for (int i = 0; i < len; i++)
        for (int j = 0; j < num.len; j++)
            result.s[i + j] += s[i] * num.s[j];

    for (int i = 0; i < result.len; i++)
    {
        result.s[i + 1] += result.s[i] / 10;
        result.s[i] %= 10;
    }
    result.clean();
    result.sign = !(sign^num.sign);
    return result;
}

bign bign::operator*(const int num)const
{
    bign x = num;
    bign z = *this;
    return x*z;
}
bign bign::operator*=(const bign&num)
{
    *this = *this * num;
    return *this;
}

bign bign::operator /(const bign&num)const
{
    bign ans;
    ans.len = len - num.len + 1;
    if (ans.len < 0)
    {
        ans.len = 1;
        return ans;
    }

    bign divisor = *this, divid = num;
    divisor.sign = divid.sign = 1;
    int k = ans.len - 1;
    int j = len - 1;
    while (k >= 0)
    {
        while (divisor.s[j] == 0) j--;
        if (k > j) k = j;
        char z[MAX_L];
        memset(z, 0, sizeof(z));
        for (int i = j; i >= k; i--)
            z[j - i] = divisor.s[i] + '0';
        bign dividend = z;
        if (dividend < divid) { k--; continue; }
        int key = 0;
        while (divid*key <= dividend) key++;
        key--;
        ans.s[k] = key;
        bign temp = divid*key;
        for (int i = 0; i < k; i++)
            temp = temp * 10;
        divisor = divisor - temp;
        k--;
    }
    ans.clean();
    ans.sign = !(sign^num.sign);
    return ans;
}

bign bign::operator/=(const bign&num)
{
    *this = *this / num;
    return *this;
}

bign bign::operator%(const bign& num)const
{
    bign a = *this, b = num;
    a.sign = b.sign = 1;
    bign result, temp = a / b*b;
    result = a - temp;
    result.sign = sign;
    return result;
}

bign bign::pow(const bign& num)const
{
    bign result = 1;
    for (bign i = 0; i < num; i++)
        result = result*(*this);
    return result;
}

bign bign::factorial()const
{
    bign result = 1;
    for (bign i = 1; i <= *this; i++)
        result *= i;
    return result;
}

void bign::clean()
{
    if (len == 0) len++;
    while (len > 1 && s[len - 1] == '\0')
        len--;
}

bign bign::Sqrt()const
{
    if(*this<0)return -1;
    if(*this<=1)return *this;
    bign l=0,r=*this,mid;
    while(r-l>1)
    {
        mid=(l+r)/2;
        if(mid*mid>*this)
            r=mid;
        else
            l=mid;
    }
    return l;
}

bign::~bign()
{
}
int main() {
	bign x,y,x1,x2,x3,y1,y2,y3,x4,y4;
	bign a,b,c,d,e,f;
	bign ling=0;
	int t;
	bign dis;
	bign tes;
	cin>>t;
	while(t--)
	{
		cin>>x1>>y1>>x2>>y2>>x3>>y3>>x4>>y4;
		x1=x1*max_n;
		y1=y1*max_n;
		x2=x2*max_n;
		y2=y2*max_n;
		y3=y3*max_n;
		x3=x3*max_n;
		x4=x4*max_n;
		y4=y4*max_n;
		a=x1-x2;
		b=y1-y2;
		c=x1-x3;
		d=y1-y3;
		e=(x1*x1-x2*x2-y2*y2+y1*y1)/2;
		f=(x1*x1-x3*x3-y3*y3+y1*y1)/2;
		x= (d*e-b*f)/(b*c-a*d);
		y= (a*f-c*e)/(b*c-a*d);
		dis=(x+x1)*(x+x1)+(y+y1)*(y+y1);
		tes=(x+x4)*(x+x4)+(y+y4)*(y+y4);
		if(dis<tes)
			printf("Accepted\n");
		else
			printf("Rejected\n");
	}
  return 0;
}

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转载自blog.csdn.net/zhouzi2018/article/details/81255101
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