2017多校第6场 HDU 6097 Mindis 计算几何，圆的反演

Posted 2020-10-02 Lsxxxxxxxxxxxxx

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6097

题意：有一个圆心在原点的圆，给定圆的半径，给定P、Q两点坐标（PO=QO，P、Q不在圆外），取圆上一点D，求PD+QD的最小值。

解法：圆的反演。

很不幸不总是中垂线上的点取到最小值，考虑点在圆上的极端情况。

做P点关于圆的反演点P‘,OPD与ODP‘相似，相似比是|OP| : r。

Q点同理。

极小化PD+QD可以转化为极小化P‘D+Q‘D。

当P‘Q‘与圆有交点时，答案为两点距离，否则最优值在中垂线上取到。

时间复杂度 O(1)

也有代数做法，结论相同。

优秀的黄金分割三分应该也是可以卡过的。

分析可以看这个博客：http://blog.csdn.net/qq_34845082/article/details/77099332

 

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
struct FastIO
{
    static const int S = 1310720;
    int wpos;
    char wbuf[S];
    FastIO() : wpos(0) {}
    inline int xchar()
    {
        static char buf[S];
        static int len = 0, pos = 0;
        if(pos == len)
            pos = 0, len = fread(buf, 1, S, stdin);
        if(pos == len)
            exit(0);
        return buf[pos ++];
    }
    inline unsigned long long xuint()
    {
        int c = xchar();
        unsigned long long x = 0;
        while(c <= 32)
            c = xchar();
        for(; ‘0‘ <= c && c <= ‘9‘; c = xchar())
            x = x * 10 + c - ‘0‘;
        return x;
    }
    inline long long xint()
    {
        long long s = 1;
        int c = xchar(), x = 0;
        while(c <= 32)
            c = xchar();
        if(c == ‘-‘)
            s = -1, c = xchar();
        for(; ‘0‘ <= c && c <= ‘9‘; c = xchar())
            x = x * 10 + c - ‘0‘;
        return x * s;
    }
    inline void xstring(char *s)
    {
        int c = xchar();
        while(c <= 32)
            c = xchar();
        for(; c > 32; c = xchar())
            * s++ = c;
        *s = 0;
    }
    inline double xdouble()
    {
        bool sign = 0;
        char ch = xchar();
        double x = 0;
        while(ch <= 32)
            ch = xchar();
        if(ch == ‘-‘)
            sign = 1, ch = xchar();
        for(; ‘0‘ <= ch && ch <= ‘9‘; ch = xchar())
            x = x * 10 + ch - ‘0‘;
        if(ch == ‘.‘)
        {
            double tmp = 1;
            ch = xchar();
            for(; ch >= ‘0‘ && ch <= ‘9‘; ch = xchar())
                tmp /= 10.0, x += tmp * (ch - ‘0‘);
        }
        if(sign)
            x = -x;
        return x;
    }
    inline void wchar(int x)
    {
        if(wpos == S)
            fwrite(wbuf, 1, S, stdout), wpos = 0;
        wbuf[wpos ++] = x;
    }
    inline void wint(long long x)
    {
        if(x < 0)
            wchar(‘-‘), x = -x;
        char s[24];
        int n = 0;
        while(x || !n)
            s[n ++] = ‘0‘ + x % 10, x /= 10;
        while(n--)
            wchar(s[n]);
    }
    inline void wstring(const char *s)
    {
        while(*s)
            wchar(*s++);
    }
    inline void wdouble(double x, int y = 8)
    {
        static long long mul[] = {1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000, 10000000000LL, 100000000000LL, 1000000000000LL, 10000000000000LL, 100000000000000LL, 1000000000000000LL, 10000000000000000LL, 100000000000000000LL};
        if(x < -1e-12)
            wchar(‘-‘), x = -x;
        x *= mul[y];
        long long x1 = (long long) floorl(x);
        if(x - floor(x) >= 0.5)
            ++x1;
        long long x2 = x1 / mul[y], x3 = x1 - x2 * mul[y];
        wint(x2);
        if(y > 0)
        {
            wchar(‘.‘);
            for(size_t i = 1; i < y && x3 * mul[i] < mul[y]; wchar(‘0‘), ++i);
            wint(x3);
        }
    }
    ~FastIO()
    {
        if(wpos)
            fwrite(wbuf, 1, wpos, stdout), wpos = 0;
    }
} io;

const double eps = 1e-8;
double getDis(double x, double y){
    return sqrt(x*x+y*y);
}
double r,x1,y1,x2,y2;

int main()
{
    int T = io.xint();
    while(T--)
    {
        r = io.xdouble();
        x1 = io.xdouble();
        y1 = io.xdouble();
        x2 = io.xdouble();
        y2 = io.xdouble();
        double d0 = getDis(x1, y1);
        if(fabs(d0)<=eps){//p和q和原点重合
            printf("%.8f\n", 2*r);
            continue;
        }
        double k = r*r/d0/d0;//用这个比例确定p和q的反演点
        double x3=x1*k, x4=x2*k;
        double y3=y1*k, y4=y2*k;
        double mx=(x3+x4)/2.0,my=(y3+y4)/2.0;
        double ans;
        double d = getDis(mx, my);
        if(d <= r){//判断反演点和半径的关系 如果两个反演点的中点到圆心的距离小于半径
            double dis = getDis(x3-x4,y3-y4);
            ans = dis*d0/r;
        }
        else{//其他的即是连线与圆相离时的状态 这时候的d点是p和q的反演点的连线的中垂线与圆的交点
            double kk=r/d;//其他的即是连线与圆相离时的状态 这时候的d点是p和q的反演点的连线的中垂线与圆的交点
            double smx = mx*kk, smy = my*kk;
            ans = 2*getDis(smx-x1,smy-y1);
        }
        printf("%.8f\n", ans);
    }
    return 0;
}