HDU 4998 Rotate --几何

题意：给n个点（x,y,p），从1~n，一次每次所有点绕着第 i 个点（原来的）逆时针转pi个弧度，问最后所有点的位置相当于绕哪个点旋转多少弧度，求出那点X和弧度P

解法：直接模拟旋转，每次计算新的坐标，最后选两个新的点分别和他们原来的点连一条线，两条线的中垂线的交点即为圆心，求出了圆心就可以求出转了多少弧度了。

注意判中垂线垂直x轴的情况以及n==1的情况。

最后角度要根据位置关系判下正负。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#define pii acos(-1.0)
#define eps 1e-8
using namespace std;
#define N 17

typedef struct point
{
   double x,y,radi;
   int ind;
   point(double x=0,double y=0):x(x),y(y){}
}Vector;

struct Point
{
    double x,y;
    Point(double x = 0,double y = 0):x(x),y(y){ }
};

Vector pi[N];
Point np[N];
int n;

int dcmp(double x)
{
    if(fabs(x)<eps)
        return 0;
    return x < 0 ? -1:1;
}
Vector operator  + (Vector A,Vector B){return Vector(A.x+B.x,A.y+B.y);}
Vector operator  -  (point A,point B){return Vector(A.x-B.x,A.y-B.y);}
Vector operator  *  (Vector A,double p){return Vector(A.x*p,A.y*p);}
Vector operator  /  (Vector A,double p){return Vector(A.x/p,A.y/p);}
bool operator ==  (const point& a,const point& b){return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;}
bool operator < (const point& a,const point& b){return a.x<b.x ||(a.x==b.x&&a.y<b.y);} //比较和排序可用
double Cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;}   //叉积 ，大于零说明B在A的左边,小于零说明B在A的右边
double Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;}    //点积
double Length(Vector A){return sqrt(Dot(A,A));}              //向量长度
double Angle(Vector A,Vector B){return acos(Dot(A,B)/Length(A)/Length(B));};
Vector Rotate(Vector A,double rad){return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}//rad为弧度，向量逆时针旋转rad

int main()
{
    int t,i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%lf%lf%lf",&pi[i].x,&pi[i].y,&pi[i].radi),pi[i].ind = i;
            np[i].x = pi[i].x;
            np[i].y = pi[i].y;
        }
        if(n == 1)
        {
            printf("%.10f %.10f %.10f\n",pi[1].x,pi[1].y,pi[1].radi);
            continue;
        }
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                Vector k = Vector(np[j].x-pi[i].x,np[j].y-pi[i].y);
                k = Rotate(k,pi[i].radi);
                np[j].x = pi[i].x + k.x;
                np[j].y = pi[i].y + k.y;
            }
        }
        Point A,B,C,D;
        A = Point(pi[1].x,pi[1].y);
        B = np[1];
        C = Point(pi[2].x,pi[2].y);
        D = np[2];
        Point Mid1 = Point((A.x+B.x)/2.0,(A.y+B.y)/2.0);
        Point Mid2 = Point((C.x+D.x)/2.0,(C.y+D.y)/2.0);
        double k1,k2;
        double Ix,Iy;
        if(dcmp(A.y-B.y) == 0)
        {
            if(dcmp(D.x-C.x) == 0)
                k2 = 0.0;
            else
            {
                k2 = (D.y-C.y)/(D.x-C.x);
                k2 = -1.0/k2;
            }
            Ix = Mid1.x;
            Iy = Mid2.y + k2*(Mid1.x-Mid2.x);
        }
        else if(dcmp(D.y-C.y) == 0)
        {
            if(dcmp(B.x-A.x) == 0)
                k1 = 0.0;
            else
            {
                k1 = (B.y-A.y)/(B.x-A.x);
                k1 = -1.0/k1;
            }
            Ix = Mid2.x;
            Iy = Mid1.y + k1*(Mid2.x-Mid1.x);
        }
        else
        {
            k1 = (B.y-A.y)/(B.x-A.x);
            k1 = -1.0/k1;
            k2 = (D.y-C.y)/(D.x-C.x);
            k2 = -1.0/k2;
            double b1 = -k1*Mid1.x + Mid1.y;
            double b2 = -k2*Mid2.x + Mid2.y;
            Ix = (b2-b1)/(k1-k2);
            Iy = k1*Ix + b1;
        }
        Vector ka = Vector(pi[1].x-Ix,pi[1].y-Iy);
        Vector kb = Vector(np[1].x-Ix,np[1].y-Iy);
        double ang = Angle(ka,kb);
        double coss = Cross(ka,kb);
        if(dcmp(coss-0.0) == -1)
            ang = 2.0*pii-ang;
        printf("%.10f %.10f %.10f\n",Ix,Iy,ang);
    }
    return 0;
}