半平面交
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半平面交
#include<bits/stdc++.h>
#define eps 1e-6
#define N 510
using namespace std;
struct Point
{
double x,y;
Point(){};
Point(double a,double b)
{
x=a,y=b;
}
}p[N],e[N];
struct Line
{
Point s,t;
double d;
Line(){};
Line(Point a,Point b)
{
s=a,t=b;
}
}l[N],q[N];
Point operator + (Point a,Point b)
{
return Point{a.x+b.x,a.y+b.y};
}
Point operator - (Point a,Point b)
{
return Point{a.x-b.x,a.y-b.y};
}
Point operator * (Point a,double x)
{
return Point{a.x*x,a.y*x};
}
int n,cnt,cnt1;
double ans;
double cross(Point a,Point b)
{
return a.x*b.y-b.x*a.y;
}
double dis(Point a,Point b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int compare(double x,double y)
{
if(fabs(x-y)<eps)return 0;
return x-y<0?-1:1;
}
bool cmp(Line a,Line b)
{
if(!compare(a.d,b.d))return cross(a.t-a.s,b.t-a.s)>0;
return a.d<b.d;
}
bool onright(Point a,Line l)
{
return compare(cross(l.t-l.s,a-l.s),0)>0?0:1;
}
Point intersection(Line a,Line b)
{
Point u=a.s-b.s;
Point v=a.t-a.s;
Point w=b.t-b.s;
double t=cross(w,u)/cross(v,w);
return a.s+v*t;
}
double angle(Point a)
{
return atan2(a.y,a.x);
}
void half()
{
sort(l+1,l+cnt+1,cmp);
cnt1=0;
for(int i=1;i<cnt;i++)
{
if(!compare(l[i+1].d-l[i].d,0))continue;
l[++cnt1]=l[i];
}
l[++cnt1]=l[cnt];
cnt=cnt1;
int L=1,R=0;
q[++R]=l[1],q[++R]=l[2];
for(int i=3;i<=cnt;i++)
{
while(L<R&&onright(intersection(q[R],q[R-1]),l[i]))R--;
while(L<R&&onright(intersection(q[L],q[L+1]),l[i]))L++;
q[++R]=l[i];
}
while(L<R&&onright(intersection(q[R],q[R-1]),q[L]))R--;
while(L<R&&onright(intersection(q[L],q[L+1]),q[R]))L++;
q[++R]=q[L];
cnt1=0;
for(int i=L;i<R;i++)e[++cnt1]=intersection(q[i],q[i+1]);
e[++cnt1]=e[1];
}
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
int m;
scanf("%d",&m);
Point last,st;
for(int j=1;j<=m;j++)
{
double x,y;
scanf("%lf%lf",&x,&y);
if(j==1)
{
last=st=Point{x,y};
continue;
}
l[++cnt]=Line{last,Point{x,y}};
last=Point{x,y};
}
l[++cnt]=Line{last,st};
}
for(int i=1;i<=cnt;i++)l[i].d=angle(l[i].s-l[i].t);
half();
if(cnt1<3)
{
puts("0.000");
return 0;
}
for(int i=1;i<cnt1;i++)ans+=cross(e[i],e[i+1]);
printf("%.3lf\n",ans/2.0);
return 0;
}