hdu 4998

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

这道题，在比赛的时候看了很久，才明白题目的大意。都怪自己不好好学习英语。后来经过队友翻译才懂是什么意思。

题目大意：二维平面内的物品，把它们绕给定的n个点，逆时针旋转。 现在求通过一个A点，逆时针旋转角度P就能完成这个操作。

问：这个点A的坐标，P的角度。

解题思路：随意找个点t1,绕着n个点旋转得到t2。点A一定在线段t1t2的垂直平分线上。再找一点t3，绕n个点旋转得到t4。

线段t1t2的垂直平分线和线段t3t4的垂直平分线相交一点，这点就是A点。证明过程就不写了，可以画图看看。

旋转的角度是向量At1和向量At2的夹角Θ，或是2∏-Θ

 #include<cstdio>
 #include<cmath>
 #include<iostream>
 using namespace std;

 struct Point{
     double x, y;

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

 typedef Point Vector;

 const double eps = 1e-;

 int dcmp(double x){
     if(fabs(x) < eps)
         return ;
     return x <  ? - : ;
 }

 bool operator == (Point A, Point B){
     return dcmp(A.x - B.x) ==  && dcmp(A.y - B.y) == ;
 }

 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){//点乘
     return A.x * B.x + A.y * B.y;
 }

 double length(Vector A){//向量的模
     return sqrt(dot(A, A));
 }

 double angle(Vector A, Vector B){//两个向量的夹角
     return acos(dot(A, B) / length(A) / length(B));
 }

 Vector rotate(Vector A, double rad){//点A绕原点旋转角度为rad
     return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
 }

 Vector normal(Vector A){//向量的法向量
     double L = length(A);
     return Vector(-A.y / L, A.x / L);
 }

 double cross(Vector A, Vector B){//叉乘
     return A.x * B.y - A.y * B.x;
 }

 int n;
 Point p[];
 double ra[];

 void input(){
     cin>> n;
     for(int i = ; i < n; ++i){
         cin>> p[i].x>> p[i].y>> ra[i];
         if(dcmp(ra[i] -  * acos(-1.0)) ==  || dcmp(ra[i]) == ){
         //当输入的弧度为0或者2π时，都是没用的
             ra[i] = ;
             n--;
             i--;
         }
     }
 }

 Vector rotate_point(Vector A){//将A点绕n个点旋转
     for(int i = ; i < n; ++i){
         A = p[i] + rotate(A - p[i], ra[i]);
     }
     return A;
 }

 Vector mid_point(Point A, Point B){//求两点之间的中点
     return Vector((A.x + B.x) / , (A.y + B.y) / );
 }

 Point get_line_intersection(Point P, Vector v, Point Q, Vector w){//两直线求交点，《算法入门经典训练之南》几何
     Vector u = P - Q;
     double t = cross(w, u) / cross(v, w);
     return P + v * t;
 }

 void solve(){
     Point t1[], t2[], mid[], vec[];
     t1[].x = -;
     t1[].y = -;
     t1[].x = -;
     t1[].y = -;
     for(int i = ; i < ; ++i){
         t2[i] = rotate_point(t1[i]);
         mid[i] = mid_point(t1[i], t2[i]);
         vec[i] = normal(t1[i] - t2[i]);
     }
     Point ans = get_line_intersection(mid[], vec[], mid[], vec[]);//答案点A
     double ansp = angle(t1[] - ans, t2[] - ans);//旋转角度P

     if(cross(t1[] - ans, t2[] - ans)  < ){//判断是ansp 还是2π - ansp
         ansp =  * acos(-1.0) - ansp;
     }

     if(dcmp(ans.x) == ){//加入答案是-1e-11，如果直接输出就会是-0.0000000000
         ans.x = ;
     }
     if(dcmp(ans.y) == ){
         ans.y = ;
     }

     printf("%.10lf %.10lf %.10lf\n", ans.x, ans.y, ansp);
 }

 int main(){
     //freopen("data.in", "r", stdin);
     //freopen("data.out", "w", stdout);
     int t;
     cin>> t;
     while(t--){
         input();
         solve();
     }
     return ;
 }

AC代码