hdu 4998 Rotate（计算几何）

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题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=4998

解题思路：

题目大意：给定n个点和n个角度，平面内任意一个点依次以这些点Point(i)为旋转中心，逆时针旋转 R(i)度，求最后相当于绕哪个点旋转多少度，求这个点的坐标以及旋转的角度。

算法思想：

一个点（x，y）绕另外一个点（x0,y0）旋转 r0 度的公式为：

x1= (x - x0)*cos(r0) - (y - y0)*sin(r0) + x0 ;

y1= (x - x0)*sin(r0) + (y - y0)*cos(r0) + y0 ;

最后旋转点肯定是xx,tmp1的垂直平分线以及由yy,tmp2的垂直平分线的交点，旋转角度rad = (r1+r2+.....+rn)%(2*pi)。

最后直接套模板即可。。。http://blog.csdn.net/piaocoder/article/details/47403265

AC代码：

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>using namespace std;const double eps = 1e-5;const double pi = acos(-1.0);struct Point{    double x,y;    Point(double x = 0,double y = 0):x(x),y(y){} // 构造函数，方便代码编写};typedef Point Vector;Vector operator + (Vector A,Vector B){    return Vector(A.x+B.x,A.y+B.y);}Vector operator - (Vector A,Vector 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);}double Dot(Vector A,Vector B){    return A.x*B.x+A.y*B.y;}double Length(Vector A){    return sqrt((Dot(A,A)));}double Cross(Vector A,Vector B){    return A.x*B.y-A.y*B.x;}Vector Rotate(Vector A,double rad){    return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}Vector Normal(Vector A){    double L = Length(A);    return Vector(-A.y/L,A.x/L);}Point GetLineIntersection(Point P,Vector v,Point Q,Vector w){    Vector u = P-Q;    double t = Cross(w,u)/Cross(v,w);    return P+v*t;}int main(){    int T;    scanf("%d",&T);    while(T--){        int n;        scanf("%d",&n);        Point xx(-1,0),yy(1,0);        Point tmp1 = xx,tmp2 = yy,tmp3;        double x, rad = 0;        for(int i = 0; i < n; i++){            scanf("%lf%lf%lf",&tmp3.x,&tmp3.y,&x);            tmp1 = Rotate(tmp1-tmp3,x)+tmp3;            tmp2 = Rotate(tmp2-tmp3,x)+tmp3;            rad += x;        }        Point tt1 = (xx+tmp1)/2, tt2 = (yy+tmp2)/2;        Vector v1,v2;        v1 = Normal(tmp1-xx);v2 = Normal(tmp2-yy);        Point ans = GetLineIntersection(tt1,v1,tt2,v2);        while(rad > 2*pi)            rad -= 2*pi;        printf("%.10lf %.10lf %.10lf\n",ans.x,ans.y,rad);    }    return 0;}

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