• 刘汝佳半平面交模板


     1 #include <algorithm>
     2 #include <iostream>
     3 #include <cstring>
     4 #include <cstdio>
     5 #include <cmath>
     6 using namespace std;
     7 #define N 50005
     8 
     9 const double eps=1e-8;
    10 int dcmp(double x) {
    11     if (x<=eps&&x>=-eps) return 0;
    12     return (x>0)?1:-1;
    13 }
    14 struct Vector {
    15     double x,y;
    16     Vector(double X=0,double Y=0){
    17         x=X,y=Y;
    18     }
    19 };
    20 typedef Vector Point;
    21 
    22 struct Line {
    23     Point p;
    24     Vector v;
    25     double ang;
    26     Line(Point P=Point(0,0),Vector V=Vector(0,0)) {
    27         p=P,v=V;
    28         ang=atan2(v.y,v.x);
    29     }
    30     bool operator < (const Line &a) const {
    31         return ang<a.ang;
    32     }
    33 };
    34 Vector operator + (Vector a,Vector b) {return Vector(a.x+b.x,a.y+b.y);}
    35 Vector operator - (Vector a,Vector b) {return Vector(a.x-b.x,a.y-b.y);}
    36 Vector operator * (Vector a,double b) {return Vector(a.x*b,a.y*b);}
    37 
    38 int n,l,r,m,cnt;
    39 double ans;
    40 Line L[N],q[N];
    41 Point p[N],poly[N];
    42 
    43 double Cross(Vector a,Vector b) {
    44     return a.x*b.y-a.y*b.x;
    45 }
    46 Point GLI(Point P,Vector v,Point Q,Vector w) {
    47     Vector u=P-Q;
    48     double t=Cross(w,u)/Cross(v,w);
    49     return P+v*t;
    50 }
    51 bool Onleft(Line m,Point P) {
    52     Vector w=P-m.p;
    53     return dcmp(Cross(m.v,w))>=0;
    54 }
    55 void halfp(){
    56     sort(L+1,L+n+1);
    57     cnt=0;
    58     q[l=r=1]=L[1];
    59     for (int i=2;i<=n;++i) {
    60         while (l<r&&!Onleft(L[i],p[r-1])) --r;
    61         while (l<r&&!Onleft(L[i],p[l])) ++l;
    62         q[++r]=L[i];
    63         if (dcmp(Cross(q[r].v,q[r-1].v))==0) {
    64             --r;
    65             if (Onleft(q[r],L[i].p))
    66                 q[r]=L[i];
    67         }
    68         if (l<r)
    69             p[r-1]=GLI(q[r-1].p,q[r-1].v,q[r].p,q[r].v);
    70     }
    71     while (l<r&&!Onleft(q[l],p[r-1]))
    72         --r;
    73     if (r-l<=1) return;
    74     p[r]=GLI(q[r].p,q[r].v,q[l].p,q[l].v);
    75     for (int i=l;i<=r;++i) poly[++cnt]=p[i];
    76 }
    77 double Area() {
    78     double ans=0;
    79     for (int i=2;i<cnt;++i)
    80         ans+=Cross(poly[i]-poly[1],poly[i+1]-poly[1]);
    81     return fabs(ans/2);
    82 }
    83 
    84 int main() {
    85     scanf("%d",&n); // 输入点的个数
    86     for (int i=1;i<=n;++i) {
    87         Point P,Q;double a,b,c,d; // 输入四个点的坐标 x1,y1,x2,y2
    88         scanf("%lf%lf%lf%lf",&a,&b,&c,&d);
    89         P=Point(a,b);Q=Point(c,d); // 得到点
    90         L[i]=Line(P,Q-P); // 通过一个点和线得到 直线
    91     } // 开始需要设置一个无限大的区域
    92     L[++n]=Line(Point(0,10000),Vector(0,-10000));
    93     L[++n]=Line(Point(0,0),Vector(10000,0));
    94     L[++n]=Line(Point(10000,0),Vector(0,10000));
    95     L[++n]=Line(Point(10000,10000),Vector(-10000,0));
    96     halfp();
    97     printf("%.1lf
    ",Area());
    98     return 0;
    99 }
    View Code

    真tm难。。。虽然挺有用的

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  • 原文地址:https://www.cnblogs.com/cmbyn/p/8860946.html
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