HDU - 5572 An Easy Physics Problem （计算几何模板）

【题目概述】

On an infinite smooth table, there's a big round fixed cylinder and a little ball whose volume can be ignored.

 

Currently the ball stands still at point A, then we'll give it an initial speed and a direction. If the ball hits the cylinder, it will bounce back with no energy losses.

 

We're just curious about whether the ball will pass point B after some time.

Input

First line contains an integer T, which indicates the number of test cases.

 

Every test case contains three lines.

 

The first line contains three integers Ox, Oy and r, indicating the center of cylinder is (Ox,Oy) and its radius is r.

 

The second line contains four integers Ax, Ay, Vx and Vy, indicating the coordinate of A is (Ax,Ay) and the initial direction vector is (Vx,Vy).

 

The last line contains two integers Bx and By, indicating the coordinate of point B is (Bx,By).

 

⋅ 1 ≤ T ≤ 100.

 

⋅ |Ox|,|Oy|≤ 1000.

 

⋅ 1 ≤ r ≤ 100.

 

⋅ |Ax|,|Ay|,|Bx|,|By|≤ 1000.

 

⋅ |Vx|,|Vy|≤ 1000.

 

⋅ Vx≠0 or Vy≠0.

 

⋅ both A and B are outside of the cylinder and they are not at same position.

Output

For every test case, you should output " Case #x: y", where x indicates the case number and counts from 1. y is " Yes" if the ball will pass point B after some time, otherwise y is " No".

Sample Input

2

0 0 1

2 2 0 1

-1 -1

0 0 1

-1 2 1 -1

1 2

Sample Output

Case #1: No

Case #2: Yes

 

【题目大意】：一个面积无限大的光滑桌面，一个质点从初始点A（x0,y0）出发，初速度为（Vx, Vy）,桌子上有一个圆柱体，圆心（cx,cy），半径为R，小球如果与圆柱体碰撞则发生的是完全弹性碰撞，在桌子上某处有一点B（ex,ey），问这个质点能不能经过B点？

 

【题解】：简单的几何学。记小球初始从A点沿着射线L1运动，分两种情况讨论。

（1）小球能和圆柱体相撞，即L1与圆O有两个交点（如果相切认为不会发生碰撞改变方向），记碰撞点为G点（射线与圆相交的较近的那一点），那么小球碰撞后沿什么射线运动呢？连接OG，则小球沿着L1关于OG的对称直线L2运动，如果B点在射线L2上则满足题意，或者在未碰撞之前，即B在线段AG上，也满足题意。

（2）小球不能与圆柱体相撞。判断B是否在射线L1上，若在则满足题意。

 

【代码】

#include <algorithm>
#include <cstdio>
#include <vector>
#include <cmath>
using namespace std;

const double eps = 1e-8; //1e-10会WA，注意调整精度，过大过小都不行

//浮点数是否等于0，在eps精度范围之内 
int dcmp(double x){
    if(fabs(x) < eps) return 0;
    return x < 0 ? -1 : 1;
}
//自定义开方 
double mySqrt(double x){
    return sqrt(max((double)0, x));
}

//点的结构体定义 
struct Point
{
    double x, y;
    Point(double x=0, double y=0):x(x), y(y){}
    Point& operator = (Point p){
        x = p.x;
        y = p.y;
        return *this;
    }
};

typedef Point Vector;
//向量四则运算定义 
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);}


//点乘定义 
double dot(Vector A, Vector B){ return A.x * B.x + A.y * B.y;}
//模长定义 
double length(Vector A){ return mySqrt(dot(A, A));}
//叉积定义 
double cross(Vector A, Vector B){ return A.x * B.y - A.y * B.x;}

//直线结构体 
struct Line
{
    //相当于点向式 
    //   (x - x0) / v.x =  (y - y0) / v.y 
    Point p;//直线上一点 
    Vector v;//方向向量定方向 
    Line(Point p, Vector v):p(p), v(v){}
    Point getPoint(double t){ 
        return Point(p.x + v.x*t, p.y + v.y*t);
    }
};

//圆的结构体定义 圆心和半径即可 
struct Circle
{
    Point c;
    double r;
    Circle(Point c, double r):c(c), r(r){}
};

//求圆和直线交点 ，返回1有两个实数解，其他返回0
int getLineCircleIntersection(Line L, Circle C, Point& P){ //返回t较小的那个点
    double a = L.v.x;
    double b = L.p.x - C.c.x;
    double c = L.v.y;
    double d = L.p.y - C.c.y;

    double e = a*a + c*c;
    double f = 2*(a*b + c*d);
    double g = b*b + d*d - C.r*C.r;

    double delta = f*f - 4*e*g;
    
    //判别式大于0 一元二次方程才有实数解 
    if(dcmp(delta) <= 0) return 0;
    //一元二次方程求根公式 
    double t = (-f - mySqrt(delta)) / (2*e);
    
    P = L.getPoint(t);
    return 1;
}

int getLineCircleIntersection2(Line L, Circle C, Point& P, Point& Q){ //返回两个点
    double a = L.v.x;
    double b = L.p.x - C.c.x;
    double c = L.v.y;
    double d = L.p.y - C.c.y;

    double e = a*a + c*c;
    double f = 2*(a*b + c*d);
    double g = b*b + d*d - C.r*C.r;

    double delta = f*f - 4*e*g;
    
    //判别式大于0 一元二次方程才有两个实数解 
    if(dcmp(delta) < 0) return 0;
    //一元二次方程求根公式 
    double t1 = (-f - mySqrt(delta)) / (2*e);
    double t2 = (-f + mySqrt(delta)) / (2*e);
    P = L.getPoint(t1);
    Q = L.getPoint(t2);
    return 1;
}
//点是否在直线上 
bool onLine(Point A, Line L){
    Vector w = A - L.p;    
    //printf("%lf %lf\n", w.x,w.y);
    //叉积为0即共线 
    if(dcmp(cross(w, L.v)) == 0)
        return true;
    else return false;
}

bool onRay(Point A, Line L){//点A在射线L(p,v)上，不含端点
    Vector w = A - L.p;
    //叉积等于0且点乘大于0 
    if(dcmp(cross(w, L.v))==0 && dcmp(dot(w, L.v)) > 0) return true;
    return false;
}

bool onSeg(Point A, Point B, Point C){//点A在线段BC上
    //叉积等于0且点乘小于0 
    return dcmp(cross(B-A, C-A))==0 && dcmp(dot(B-A, C-A))<0;
}

Point project(Point A, Line L){
    return L.p + L.v * ( dot(L.v, A - L.p) / dot(L.v, L.v) );
}

//求点A关于直线L的对称点 
Point mirrorPoint(Point A, Line L){
    Vector D = project(A, L);
    //printf("project: %lf, %lf\n", D.x, D.y);
    return D + (D - A);
}

int main()
{
    int T;
    int ans = 0;
    double R;
    Point O, A, B;
    Vector V;
    /*
    A = Point(1,0);
    V = Point(1,1);
    Line L1 =  Line(A,V);
    
    O.x = 2;
    O.y = 0;
    
    Circle yuanO = Circle(O, 1);
    int tt = getLineCircleIntersection2( L1,  yuanO, A, B);
    printf("%lf %lf\n",A.x,A.y);
    printf("%lf %lf\n",B.x,B.y);
    */
   /* A = Point(1,0);
    V = Point(1,1);
    Line l1 =  Line(A,V);
    printf("%lf %lf %lf %lf\n",l1.p.x,l1.p.y,l1.v.x,l1.v.y);
    B = Point();
    while(scanf("%lf%lf", &B.x, &B.y) != EOF){
        printf("%lf %lf\n",B.x,B.y);
        printf("%d\n",onLine( B,  l1));
    }*/
    
    
    
    scanf("%d", &T);
    for(int ca = 1; ca <= T; ca++){
        
        scanf("%lf%lf%lf", &O.x, &O.y, &R);
        scanf("%lf%lf%lf%lf", &A.x, &A.y, &V.x, &V.y);
        scanf("%lf%lf", &B.x, &B.y);
        
        Line LA = Line(A, V);
        Circle yuanO = Circle(O, R);
        Point C;
        
        if(getLineCircleIntersection(LA, yuanO, C)){
            if(onSeg(B, A, C)) ans = 1;
            else{
                //直线OC是对称轴 
                Line OC = Line(O, Vector(C.x - O.x, C.y - O.y));
                Point A1 = mirrorPoint(A, OC);
                // printf("%lf, %lf\n", C.x, C.y);
                // printf("%lf, %lf\n", A1.x, A1.y);
                
                //射线CB 
                Line CB = Line(C, Vector(B.x - C.x, B.y - C.y));
                
                if(onRay(A1, CB)){
                     ans = 1;

                }
                else ans = 0;
            }
        }else{
            if(onRay(B, LA)) ans = 1;
            else ans = 0;
        }
        printf("Case #%d: %s\n", ca, ans ? "Yes" : "No");
    }

    return 0;
} 

 

判断线段是否相交，点是否在矩形内

//判断一个点是否在矩形内部
#include<cstdio>
#include<iostream>
using namespace std;

typedef struct Point
{
    double x;
    double y;
    Point(){
    }
    
    Point(double x,double y)
    {
        this->x = x;
        this->y = y;
    }
}point;
// 计算 |p1 p2| X |p1 p|
double GetCross(point p1,point p2,point p)
{
    return (p2.x - p1.x) * (p.y - p1.y) -(p.x - p1.x) * (p2.y - p1.y);
}

//判断点是否在正方形内（便于测试）
bool IsPointInMatrix(point p, point p1, point p2, point p3, point p4)
{
    return GetCross(p1,p2,p) * GetCross(p3,p4,p) >= 0 && GetCross(p2,p3,p) * GetCross(p4,p1,p) >= 0;
    //return false;
}
  
double Cross_Prouct(point A,point B,point C)       //  计算BA叉乘CA；  
{  
    return (B.x-A.x)*(C.y-A.y)-(B.y-A.y)*(C.x-A.x);  
}  

bool Intersect(point A,point B,point C,point D)     //  通过叉乘判断线段是否相交；  
{  
    if(min(A.x,B.x)<=max(C.x,D.x)&&         //  快速排斥实验；  
       min(C.x,D.x)<=max(A.x,B.x)&&  
       min(A.y,B.y)<=max(C.y,D.y)&&  
       min(C.y,D.y)<=max(A.y,B.y)&&  
       Cross_Prouct(A,B,C)*Cross_Prouct(A,B,D)<=0&&      //  跨立实验；  
       Cross_Prouct(C,D,A)*Cross_Prouct(C,D,B)<=0)       //  叉乘异号表示在两侧；  
       return true;  
    else return false;  
}


int main()
{
    bool f1 = 1, f2 = 1 , flag = 0;
    
    point pset1[4];
    point pset2[4];
    for(int i=0; i<4; i++)    cin>>pset1[i].x>>pset1[i].y;
    for(int i=0; i<4; i++)    cin>>pset2[i].x>>pset2[i].y;
    
    
    for(int i=0; i<4; i++){
        bool b = IsPointInMatrix(pset1[i], pset2[0], pset2[1], pset2[2], pset2[3]);
        //cout<<b<<endl;
        if(!b){
            f1 = 0;
        }
            
    }
    
    for(int i=0; i<4; i++){
        bool b = IsPointInMatrix(pset2[i], pset1[0], pset1[1], pset1[2], pset1[3]);
        //cout<<b<<endl;
        if(!b){
            f2 = 0;
        }        
    }
    
    if(f1 || f2){
        cout<<"YES"<<endl;
        return 0;
    }
    
    //cout<<11111111111111<<endl;
    for(int i=0; i<4; i++){
        for(int j=0; j<4; j++){
        
            bool b = Intersect(pset1[i], pset1[(i+1)%4],pset2[j],pset2[(j+1)%4]);
            //cout<<b<<endl;
            
            if(b){
                flag = 1;
                break;
            }
        }
        if(flag)    break;
    }
    if(flag)    cout<<"YES"<<endl;
    else    cout<<"NO"<<endl;
    
    
    return 0;
}

 

转载于:https://www.cnblogs.com/czsharecode/p/9601847.html