求出所有交点后从左往右扫描线，用每段的中位线去截所有三角形，算出长度并后乘以该段长度即可，时间复杂度$O(n^3\log n)$。

 

#include
     
      #include
      
       #include
       
        using namespace std;const int N=310;const double eps=1e-9,inf=2000000;struct P{  double x,y;  P(){x=y=0;}  P(double _x,double _y){x=_x,y=_y;}  P operator+(P v){return P(x+v.x,y+v.y);}  P operator-(P v){return P(x-v.x,y-v.y);}  P operator*(double v){return P(x*v,y*v);}  P operator/(double v){return P(x/v,y/v);}  double operator*(P v){return x*v.x+y*v.y;}}tri[N][4],seg[N];inline bool cmp(P a,P b){return a.x
        
         0?1:-1;}inline double cross(P a,P b){return a.x*b.y-a.y*b.x;}inline bool has_intersection(P a,P b,P p,P q){  int d1=sig(cross(b-a,p-a)),d2=sig(cross(b-a,q-a)),      d3=sig(cross(q-p,a-p)),d4=sig(cross(q-p,b-p));  return d1*d2<0&&d3*d4<0;}inline P line_intersection(P a,P b,P p,P q){  double U=cross(p-a,q-p),D=cross(b-a,q-p);  return a+(b-a)*(U/D);}inline double cal(double x){  P D(x,-inf),U(x,inf);  int i,m=0;  for(i=0;i
         
          1)sort(seg,seg+m,cmp);  double l=-inf,r=-inf,t=0;  for(i=0;i
          
           0)t+=r-l,l=seg[i].x; r=max(r,seg[i].y); } return t+r-l;}int main(){ scanf("%d",&n); for(i=0;i