poj 2451 Uyuw's Concert（半平面交）

 1 #include<cmath>
 2 #include<cstdio>
 3 #include<vector>
 4 #include<cstring>
 5 #include<algorithm>
 6 #define FOR(a,b,c) for(int a=(b);a<=(c);a++)
 7 using namespace std;
 8 
 9 const int eps = 1e-10;
10 
11 struct Pt {
12     double x,y;
13     Pt (double x=0,double y=0) :x(x),y(y) {}
14 };
15 typedef Pt vec;
16 
17 vec operator - (Pt a,Pt b) { return vec(a.x-b.x,a.y-b.y); }
18 vec operator + (vec a,vec b) { return vec(a.x+b.x,a.y+b.y); }
19 vec operator * (vec a,double x) { return vec(a.x*x,a.y*x); }
20 
21 double cross(Pt a,Pt b) { return a.x*b.y-a.y*b.x; }
22 
23 struct Line {
24     Pt p; vec v; double ang;
25     Line() {}
26     Line(Pt p,vec v) :p(p),v(v) { ang=atan2(v.y,v.x); }
27     bool operator < (const Line& rhs) const {
28         return ang < rhs.ang;
29     }
30 };
31 //p在l的左边
32 bool onleft(Line L,Pt p) { return cross(L.v,p-L.p)>0; }
33 Pt getLineInter(Line a,Line b) {
34     vec u=a.p-b.p;
35     double t=cross(b.v,u)/cross(a.v,b.v);
36     return a.p+a.v*t;
37 }
38 
39 vector<Pt> HPI(vector<Line> L) {
40     int n=L.size();
41     sort(L.begin(),L.end());
42     int f,r;
43     vector<Pt> p(n) , ans;
44     vector<Line> q(n);
45     q[f=r=0]=L[0];
46     for(int i=1;i<n;i++) {
47         while(f<r && !onleft(L[i],p[r-1])) r--;
48         while(f<r && !onleft(L[i],p[f])) f++;
49         q[++r]=L[i];
50         if(fabs(cross(q[r].v,q[r-1].v))<eps) {
51             r--;
52             if(onleft(q[r],L[i].p)) q[r]=L[i];
53         }
54         if(f<r) p[r-1]=getLineInter(q[r-1],q[r]);
55     }
56     while(f<r && !onleft(q[f],p[r-1])) r--;
57     if(r-f<=1) return ans;
58     p[r]=getLineInter(q[r],q[f]);
59     for(int i=f;i<=r;i++) ans.push_back(p[i]);
60     return ans;
61 }
62 vector<Line> L;
63 vector<Pt> p;
64 int n;
65 
66 int main() {
67     //freopen("in.in","r",stdin);
68     //freopen("out.out","w",stdout);
69     scanf("%d",&n);
70     double x1,y1,x2,y2;
71     for(int i=0;i<n;i++) {
72         scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
73         Pt a(x1,y1) , b(x2,y2);
74         L.push_back(Line(a,b-a));
75     }
76     Pt a(0,0),b(10000,0),c(10000,10000),d(0,10000);
77     L.push_back(Line(a,b-a));
78     L.push_back(Line(b,c-b));
79     L.push_back(Line(c,d-c));
80     L.push_back(Line(d,a-d));
81     p = HPI(L);
82     double ans=0; int m=p.size();
83     for(int i=1;i<m-1;i++)
84         ans += cross(p[i]-p[0],p[i+1]-p[0]);
85     printf("%.1lf",ans/2);
86     return 0;
87 }