多角形面积相交模板

#include<cstdio> 
#include<iostream> 
#include<algorithm> 
#include<cstring> 
#include<cmath> 
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--; 
} 
//---------------华丽的分隔线-----------------// 
//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;//assumeresispositive! 
} 
//hdu-3060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; 
}

转自：https://www.cnblogs.com/ww123/p/10726572.html