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;
}