hdu3060Area2(任意多边形相交面积)

链接

多边形的面积求解是通过选取一个点（通常为原点或者多边形的第一个点）和其它边组成的三角形的有向面积。

对于两个多边形的相交面积就可以通过把多边形分解为三角形，求出三角形的有向面积递加。三角形为凸多边形，因此可以直接用凸多边形相交求面积的模板。

凸多边形相交后的部分肯定还是凸多边形，所以只需要判断哪些点是相交部分上的点，最后求下面积。

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 510
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 struct point
 {
     double x,y;
     point(double x=,double y=):x(x),y(y) {} //构造函数 方便代码编写
 }p[N],q[N];
 typedef point pointt;
 pointt operator + (point a,point b)
 {
     return point(a.x+b.x,a.y+b.y);
 }
 pointt operator - (point a,point b)
 {
     return point(a.x-b.x,a.y-b.y);
 }
 int dcmp(double x)
 {
     if(fabs(x)<eps) return ;
     else return x<?-:;
 }
 bool operator == (const point &a,const point &b)
 {
     return dcmp(a.x-b.x)==&&dcmp(a.y-b.y)==;
 }
 double cross(point a,point b)
 {
     return a.x*b.y-a.y*b.x;
 }
 double Polyarea(point p[],int n)
 {
     if(n<) return ;
     double area = ;
     for(int i =  ; i < n ; i++)
     area+=cross(p[i],p[i+]);
     return fabs(area)/;
 }
 //double Polyarea(point p[], int n)
 //{
 //    if(n < 3) return 0.0;
 //    double s = p[0].y * (p[n - 1].x - p[1].x);
 //    p[n] = p[0];
 //    for(int i = 1; i < n; ++ i)
 //        s += p[i].y * (p[i - 1].x - p[i + 1].x);
 //    return fabs(s * 0.5);
 //}
 bool intersection1(point p1, point p2, point p3, point p4, point& p)      // 直线相交
 {
     double a1, b1, c1, a2, b2, c2, d;
     a1 = p1.y - p2.y;
     b1 = p2.x - p1.x;
     c1 = p1.x*p2.y - p2.x*p1.y;
     a2 = p3.y - p4.y;
     b2 = p4.x - p3.x;
     c2 = p3.x*p4.y - p4.x*p3.y;
     d = a1*b2 - a2*b1;
     if (!dcmp(d))    return false;
     p.x = (-c1*b2 + c2*b1) / d;
     p.y = (-a1*c2 + a2*c1) / d;
     return true;
 }

 double convexpolygon(point p[],point q[],int n,int m)
 {
     int i,j;
     point ch[],cnt[];
     p[n] = p[],q[m] = q[];
     memcpy(ch,q,sizeof(point)*(m+));
     int f2,g =  ;
     for(i =  ;i < n; i++)
     {
         int f1 = dcmp(cross(p[i+]-p[i],ch[]-p[i]));
         g =  ;
         for(j =  ;j < m; j++,f1 = f2)
         {
             if(f1>=) cnt[g++] = ch[j];
             f2 = dcmp(cross(p[i+]-p[i],ch[j+]-p[i]));
             if((f1^f2)==-)
             {
                 point pp;
                 intersection1(p[i],p[i+],ch[j],ch[j+],pp);
                 cnt[g++] = pp;
             }
         }
         cnt[g] = cnt[];
         memcpy(ch,cnt,sizeof(point)*(g+));
         m = g;
     }
     return Polyarea(ch,g);
 }
 double solve(point p[],point q[],int n,int m)
 {
     int i,j;
     double area = ;
     point tp[],tq[];
     tp[] = p[];tq[] = q[];
     for(i =  ;i < n-; i++)
     {
         int k1 = dcmp(cross(p[i]-p[],p[i+]-p[]));
         tp[] = p[i],tp[] = p[i+];
         if(k1 < ) swap(tp[],tp[]);

         for(j =  ;j < m- ;j++)
         {
             tq[] = q[j],tq[] = q[j+];
             int k2 = dcmp(cross(q[j]-q[],q[j+]-q[]));
             if(k2<) swap(tq[],tq[]);

             area+=convexpolygon(tp,tq,,)*k1*k2;
         }
     }
     return Polyarea(p,n)+Polyarea(q,m)-area;
 }
 int main()
 {
     int n,m,i;
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         for(i = ; i < n ;i++)
         {
             scanf("%lf%lf",&p[i].x,&p[i].y);
         }
         for(i = ; i < m; i++)
         {
             scanf("%lf%lf",&q[i].x,&q[i].y);
         }
         p[n] = p[],q[m] = q[];
         double ans = solve(p,q,n,m);
         printf("%.2f\n",ans);
     }
     return ;
 }