两个任意多边形的交并面积

#include<bits/stdc++.h>
using namespace std;
#define maxn 510
const double eps=1E-8;
int sig(double d){
return(d>eps)-(d<-eps);
}
struct Point{
double x,y; Point(){}
Point(double x,double y):x(x),y(y){}
bool operator==(const Point&p)const{
return sig(x-p.x)==0&&sig(y-p.y)==0;
}
};
double cross(Point o,Point a,Point b){
return(a.x-o.x)*(b.y-o.y)-(b.x-o.x)*(a.y-o.y);
}
double area(Point* ps,int n){
ps[n]=ps[0];
double res=0;
for(int i=0;i<n;i++){
res+=ps[i].x*ps[i+1].y-ps[i].y*ps[i+1].x;
}
return res/2.0;
}
int lineCross(Point a,Point b,Point c,Point d,Point&p){
double s1,s2;
s1=cross(a,b,c);
s2=cross(a,b,d);
if(sig(s1)==0&&sig(s2)==0) return 2;
if(sig(s2-s1)==0) return 0;
p.x=(c.x*s2-d.x*s1)/(s2-s1);
p.y=(c.y*s2-d.y*s1)/(s2-s1);
return 1;
}

//多边形切割
//用直线ab切割多边形p，切割后的在向量(a,b)的左侧，并原地保存切割结果
//如果退化为一个点，也会返回去,此时n为1
void polygon_cut(Point*p,int&n,Point a,Point b){
static Point pp[maxn];
int m=0;p[n]=p[0];
for(int i=0;i<n;i++){
if(sig(cross(a,b,p[i]))>0) pp[m++]=p[i];
if(sig(cross(a,b,p[i]))!=sig(cross(a,b,p[i+1])))
lineCross(a,b,p[i],p[i+1],pp[m++]);
}
n=0;
for(int i=0;i<m;i++)
if(!i||!(pp[i]==pp[i-1]))
p[n++]=pp[i];
while(n>1&&p[n-1]==p[0])n--;
}

//返回三角形oab和三角形ocd的有向交面积,o是原点//
double intersectArea(Point a,Point b,Point c,Point d){
Point o(0,0);
int s1=sig(cross(o,a,b));
int s2=sig(cross(o,c,d));
if(s1==0||s2==0)return 0.0;//退化，面积为0
if(s1==-1) swap(a,b);
if(s2==-1) swap(c,d);
Point p[10]={o,a,b};
int n=3;
polygon_cut(p,n,o,c);
polygon_cut(p,n,c,d);
polygon_cut(p,n,d,o);
double res=fabs(area(p,n));
if(s1*s2==-1) res=-res;return res;
}

//求两多边形的交面积
double intersectArea(Point*ps1,int n1,Point*ps2,int n2){
if(area(ps1,n1)<0) reverse(ps1,ps1+n1);
if(area(ps2,n2)<0) reverse(ps2,ps2+n2);
ps1[n1]=ps1[0];
ps2[n2]=ps2[0];
double res=0;
for(int i=0;i<n1;i++){
for(int j=0;j<n2;j++){
res+=intersectArea(ps1[i],ps1[i+1],ps2[j],ps2[j+1]);
}
}
return res;
}

//求两个任意简单多边形的并面积
Point ps1[maxn],ps2[maxn];
int n1,n2;
int main()
{
while(scanf("%d%d",&n1,&n2)!=EOF){
for(int i=0;i<n1;i++)
scanf("%lf%lf",&ps1[i].x,&ps1[i].y);
for(int i=0;i<n2;i++)
scanf("%lf%lf",&ps2[i].x,&ps2[i].y);
double ans=intersectArea(ps1,n1,ps2,n2);
ans=fabs(area(ps1,n1))+fabs(area(ps2,n2))-ans;//容斥
printf("%.2f\n",ans);
}
return 0;
}