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 }
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真tm难。。。虽然挺有用的