[HDU4316]Mission Impossible（计算几何/凸包/半平面交）

Posted 2022-05-20 lanrtabe

题目链接：

HDU4316

计算几何太duliu了~

题目大意：空间中有一个物体，上空有\(3\)个摄像头拍摄地面，求摄像头公共盲区面积大小

首先求出每一个摄像头的盲区：

将物体的每一个点投影到地面再求凸包即可。

然后将\(3\)个凸包的每一条边都拿来一起做一次半平面交，求出公共盲区

最后三角剖分求面积就好了。

代码：

#include <cmath>
#include <cstdio>
#include <vector>
#include <algorithm>

const double Eps=1e-8;
inline double Abs(double x)return x>=0?x:-x;
inline int Sign(double x)return Abs(x)<Eps?0:(x>0?1:-1);

typedef struct Point

    double x,y;
    inline Point operator+(Point o)return (Point)x+o.x,y+o.y;
    inline Point operator-(Point o)return (Point)x-o.x,y-o.y;
    inline Point operator*(double k)return (Point)x*k,y*k;
    inline bool operator<(Point o)return Sign(x-o.x)?x<o.x:y<o.y;
    inline bool operator==(Point o)return !Sign(x-o.x)&&!Sign(y-o.y);
Vector;
inline double Dot(Vector A,Vector B)return A.x*B.x+A.y*B.y;
inline double Cross(Vector A,Vector B)return A.x*B.y-A.y*B.x;

struct Point_3Ddouble x,y,z;;
inline Point Projection(Point Light,Point_3D Object)//Object在Light下的投影
return Light+((Vector)Object.x,Object.y-Light)*(100/(100-Object.z));

struct Line

    Point p;Vector v;double Ang;Line()
    Line(Point ps,Vector vs):p(ps),v(vs)Ang=atan2(v.y,v.x);
    inline bool operator<(Line o)return Ang<o.Ang;
;

Point_3D Mac[105];//机器坐标
Point Camera[5];//摄像头坐标

Point S[105];
std::vector<Point> Convex_Hull(std::vector<Point> p)//求凸包

    std::vector<Point> Ans;
    std::sort(p.begin(),p.end());
    int n=std::unique(p.begin(),p.end())-p.begin(),Sn=0;
    for(int i=0;i<=n-1;S[++Sn]=p[i++])
        while(Sn>=2&&Sign(Cross(p[i]-S[Sn-1],S[Sn]-S[Sn-1]))<0)--Sn;
    for(int i=1;i<Sn;++i)Ans.push_back(S[i]);
    Sn=0;
    for(int i=n-1;i>=0;S[++Sn]=p[i--])
        while(Sn>=2&&Sign(Cross(p[i]-S[Sn-1],S[Sn]-S[Sn-1]))<0)--Sn;
    for(int i=1;i<Sn;++i)Ans.push_back(S[i]);
    return Ans;


inline bool OnLeft(Point p,Line l)return Sign(Cross(p-l.p,l.v))==-1;//点p是否在线l左侧
inline Point CrossP(Line A,Line B)return A.p+A.v*(Cross(B.v,A.p-B.p)/Cross(A.v,B.v));//直线交点

Point P[305];
Line Q[305];
std::vector<Point> HPI(std::vector<Line> L)//半平面交

    std::sort(L.begin(),L.end());
    int n=L.size(),Qh=0,Qt=0;
    Q[0]=L[0];
    for(int i=1;i<n;++i)
    
        while(Qh<Qt&&!OnLeft(P[Qt-1],L[i]))--Qt;
        while(Qh<Qt&&!OnLeft(P[Qh],L[i]))++Qh;
        if(Sign(Cross(L[i].v,Q[Qt].v)))Q[++Qt]=L[i];
        else if(OnLeft(L[i].p,Q[Qt]))Q[Qt]=L[i];
        if(Qh<Qt)P[Qt-1]=CrossP(Q[Qt-1],Q[Qt]);
    
    while(Qh<Qt&&!OnLeft(P[Qt-1],Q[Qh]))--Qt;
    std::vector<Point> Ans;
    if(Qt-Qh<=1)return Ans;//没有公共区域
    P[Qt]=CrossP(Q[Qh],Q[Qt]);
    for(int i=Qh;i<=Qt;++i)Ans.push_back(P[i]);
    return Ans;


double Area(std::vector<Point> Poly)//求多边形面积

    double Sum=0;
    int n=Poly.size();
    for(int i=0;i<n;++i)Sum+=Cross(Poly[i],Poly[(i+1)%n]);
    return Sum/2;


int main()

    //freopen("in.txt","r",stdin);
    for(int n;~scanf("%d",&n);)
    
        for(int i=1;i<=n;++i)scanf("%lf%lf%lf",&Mac[i].x,&Mac[i].y,&Mac[i].z);
        for(int i=1;i<=3;++i)scanf("%lf%lf",&Camera[i].x,&Camera[i].y);
        std::vector<Point> Shadow[4];
        std::vector<Line> Ls;
        for(int i=1;i<=3;++i)
        
            for(int j=1;j<=n;++j)Shadow[i].push_back(Projection(Camera[i],Mac[j]));//投影到地面
            Shadow[i]=Convex_Hull(Shadow[i]);//求出凸包
            int Ss=Shadow[i].size();
            for(int j=0;j<Ss;++j)Ls.push_back((Line)Shadow[i][(j+1)%Ss],Shadow[i][j]-Shadow[i][(j+1)%Ss]);//转化位半平面交问题
        
        Shadow[0]=HPI(Ls);
        printf("%.2f\n",Abs(Area(Shadow[0])));//面积取绝对值
    
    return 0;