HDU 5572 An Easy Physics Problem (计算几何 点类 线类 向量类 线段与圆相交)

题目链接

2015上海区域赛现场赛第5题 HDU5572

题目大意

光滑平面上，有一个固定的圆，圆外有两点A,B，问A以速度矢量V运动能否碰到B。(A碰到圆后会发生弹性碰撞)

分析

写这道题的时候没有想清楚就开始写了，比赛的时候一定要先把思路理清楚，情况比较多的题要画画树状图或者流程图。
看了一个大佬的博客中画了一个清晰的流程图，借鉴一下。

原博客链接

代码

#include<cstdio>#include<iostream>#include<cmath>#include<cstring>#include<cstdlib>#include<queue>#include<map>#include<algorithm>#include<set>#include<stack>using namespace std;typedef long long int LL;typedef long double LD;const long double eps=1e-8;int dcmp(long double x){    if (fabs(x)<eps) return 0;    if (x>0) return 1;    return -1;}long double mysqrt(long double x){    return sqrt(max((long double)0,x));}struct Point ///点类{    long double x,y;    Point() {}    Point(long double a,long double b):x(a),y(b){}    void input()    {        cin>>x>>y;    }    friend bool operator == (const Point &a,const Point &b)    {        return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0;    }};typedef Point Vector;///向量类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,long double t) {return Vector(A.x*t,A.y*t);}Vector operator / (Vector A,long double t) {return Vector(A.x/t,A.y/t);}long double dot(Vector A,Vector B) {return A.x*B.x+A.y*B.y;}long double cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;}long double length(Vector A) {return mysqrt(dot(A,A));}/*long double angle(Vector A,Vector B) {return acos(dot(A,B)/length(A)/length(B));}///求两向量夹角Vector Rotate(Vector A,long double rad)///向量A绕起点逆时针旋转rad弧度{    return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}*/struct Line///线类{    Point p;    Vector v;    Line (Point p,Vector v):p(p),v(v){}    Point getPoint(long double t)///得到线上的一点    {        return Point(p.x+t*v.x,p.y+t*v.y);    }};Point project(Point A,Line L){    return L.p+L.v*(dot(L.v,A-L.p)/dot(L.v,L.v));}Point mirrorPoint(Point A,Line L){    Vector D=project(A,L);    return D+(D-A);}void circle_cross_line(Point a,Point b,Point o,long double r,Point ret[],int &num)///求线段与圆交点模板{    long double x0=o.x,y0=o.y;    long double x1=a.x,y1=a.y;    long double x2=b.x,y2=b.y;    long double dx=x2-x1,dy=y2-y1;    long double A=dx*dx+dy*dy;    long double B=2*dx*(x1-x0)+2*dy*(y1-y0);    long double C=(x1-x0)*(x1-x0)+(y1-y0)*(y1-y0)-r*r;    long double delta=B*B-4*A*C;    num=0;    if (dcmp(delta)>=0)    {        long double t1=(-B-mysqrt(delta))/(2*A);        long double t2=(-B+mysqrt(delta))/(2*A);        if (dcmp(t1)>=0)            ret[++num]=Point(x1+t1*dx,y1+t1*dy);        if (dcmp(t2)>=0)            ret[++num]=Point(x1+t2*dx,y1+t2*dy);    }}int get_dis(Point a,Point b){    return ( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}bool onRay(Point A,Line L)///判断点A是否在射线L(p,v)上{    Vector w=A-L.p;    return (dcmp(cross(w,L.v))==0&&dcmp(dot(w,L.v))>0);}bool onSeg(Point A,Point B,Point C)///判断点A是否在线段BC上{    return (dcmp(cross(B-A,C-A))==0 && dcmp(dot(B-A,C-A))<0);}int main(){    int T,temp1,temp2,tot;    Point O,A,AA,A1,B,C,ret[5],p1,p2,p3;    Vector V;    long double r;    scanf("%d",&T);    for (int tt=1;tt<=T;tt++)    {        printf("Case #%d: ",tt);        O.input();cin>>r;        A.input();V.input();        Line LA=Line(A,V);        AA=LA.getPoint(1.0);        B.input();        circle_cross_line(A,AA,O,r,ret,tot);        if (tot==2&&ret[1]==ret[2])///用这个模板相同的交点会存两次            tot--;        if (tot<=1)///如果圆没有改变A的运动轨迹(射线与圆相离或相切)        {            if (onRay(B,LA))                printf("Yes\n");            else                printf("No\n");            continue;        }        temp1=get_dis(A,ret[1]);        temp2=get_dis(A,ret[2]);        if (temp1<temp2)///离A最近的那个交点为碰撞点C            C=ret[1];        else            C=ret[2];        if (onSeg(B,A,C))///如果B在线段AC上            printf("Yes\n");        else        {            Line OC=Line(O,C-O);            A1=mirrorPoint(A,OC);///做A关于OC的对称点AC            if (onRay(B,Line(C,A1-C)))///判断B是否在射线CA1上                printf("Yes\n");            else                printf("No\n");        }    }    return 0;}