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

typedef Point Vector;

const double eps = 1e-10;

int dcmp(double x){
    if(fabs(x) < eps)
        return 0;
    return x < 0 ? -1 : 1;
}

bool operator == (Point A, Point B){
    return dcmp(A.x - B.x) == 0 && dcmp(A.y - B.y) == 0;
}

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[11];
double ra[11];

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

Vector rotate_point(Vector A){//将A点绕n个点旋转
    for(int i = 0; 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) / 2, (A.y + B.y) / 2);
}

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[2], t2[2], mid[2], vec[2];
    t1[0].x = -10;
    t1[0].y = -10;
    t1[1].x = -100;
    t1[1].y = -20;
    for(int i = 0; i < 2; ++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[0], vec[0], mid[1], vec[1]);//答案点A
    double ansp = angle(t1[0] - ans, t2[0] - ans);//旋转角度P

    if(cross(t1[0] - ans, t2[0] - ans)  < 0){//判断是ansp 还是2π - ansp
        ansp = 2 * acos(-1.0) - ansp;
    }

    if(dcmp(ans.x) == 0){//加入答案是-1e-11，如果直接输出就会是-0.0000000000
        ans.x = 0;
    }
    if(dcmp(ans.y) == 0){
        ans.y = 0;
    }

    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 0;
}