HDU1632+半平面交

模板题

题意:给定两个凸多边形,求出合并后的面积,这个合并后的面积不包括重叠部分。

  1 #include<stdio.h>
  2 #include<string.h>
  3 #include<stdlib.h>
  4 #include<math.h>
  5 #include<algorithm>
  6 using namespace std;
  7 const int maxn = 155;
  8 const int maxm = 155;
  9 const double eps = 1e-8;
 10 const double pi = acos(-1.0);
 11 struct Point{
 12     double x,y;
 13 };
 14 struct Line{
 15     Point a,b;
 16 };
 17 Point pnt1[ maxn ],res[ maxm ],pnt2[ maxn ],tp[ maxm ];
 18 double xmult( Point op,Point sp,Point ep ){
 19     return (sp.x-op.x)*(ep.y-op.y)-(sp.y-op.y)*(ep.x-op.x);
 20 }
 21 double dist( Point a,Point b ){
 22     return sqrt( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y) );
 23 }
 24 void Get_equation( Point p1,Point p2,double &a,double &b,double &c ){
 25     a = p2.y-p1.y;
 26     b = p1.x-p2.x;
 27     c = p2.x*p1.y-p1.x*p2.y;
 28 }//直线方程
 29 Point Intersection( Point p1,Point p2,double a,double b,double c ){
 30     double u = fabs( a*p1.x+b*p1.y+c );
 31     double v = fabs( a*p2.x+b*p2.y+c );
 32     Point tt;
 33     tt.x = (p1.x*v+p2.x*u)/(u+v);
 34     tt.y = (p1.y*v+p2.y*u)/(u+v);
 35     return tt;
 36 }//交点、按照三角比例求出交点
 37 double GetArea( Point p[],int n ){
 38     double sum = 0;
 39     for( int i=2;i<n;i++ ){
 40         sum += xmult( p[1],p[i],p[i+1] );
 41     }
 42     return -sum/2.0;
 43 }//面积,顺时针为正
 44 void cut( double a,double b,double c,int &cnt ){
 45     int temp = 0;
 46     for( int i=1;i<=cnt;i++ ){
 47         if( a*res[i].x+b*res[i].y+c>-eps ){//>=0
 48             tp[ ++temp ] = res[i];
 49         }
 50         else{
 51             if( a*res[i-1].x+b*res[i-1].y+c>eps ){
 52                 tp[ ++temp ] = Intersection( res[i-1],res[i],a,b,c );
 53             }
 54             if( a*res[i+1].x+b*res[i+1].y+c>eps ){
 55                 tp[ ++temp ] = Intersection( res[i],res[i+1],a,b,c );
 56             }
 57         }
 58     }
 59     for( int i=1;i<=temp;i++ )
 60         res[i] = tp[i];
 61     res[ 0 ] = res[ temp ];
 62     res[ temp+1 ] = res[ 1 ];
 63     cnt = temp;
 64 }
 65 
 66 int main(){
 67     int m,n;
 68     while( scanf("%d",&n)==1,n ){
 69         for( int i=1;i<=n;i++ ){
 70             scanf("%lf%lf",&pnt1[i].x,&pnt1[i].y);
 71         }
 72         scanf("%d",&m);
 73         for( int i=1;i<=m;i++ ){
 74             scanf("%lf%lf",&pnt2[i].x,&pnt2[i].y);
 75         }
 76         double sumArea1,sumArea2,Area;
 77         sumArea1 = GetArea( pnt1,n );
 78         sumArea2 = GetArea( pnt2,m );
 79         if( sumArea1<eps ){
 80             reverse( pnt1+1,pnt1+1+n );
 81         }
 82         pnt1[ 0 ] = pnt1[ n ];
 83         pnt1[ n+1 ] = pnt1[ 1 ];
 84         if( sumArea2<eps ){
 85             reverse( pnt2+1,pnt2+1+m );
 86         }
 87         pnt2[ 0 ] = pnt2[ m ];
 88         pnt2[ m+1 ] = pnt2[ 1 ];
 89         for( int i=0;i<=n+1;i++ ){
 90             res[i] = pnt1[i];
 91         }
 92         int cnt = n;
 93         for( int i=1;i<=m;i++ ){
 94             double a,b,c;  
 95             Get_equation( pnt2[i],pnt2[i+1],a,b,c );  
 96             cut(a,b,c,cnt);  
 97         }  
 98         Area = GetArea( res,cnt );
 99         double ans = fabs(sumArea1)+fabs(sumArea2)-2.0*fabs(Area);
100         printf("%8.2lf",ans);
101     }
102     puts("");
103     return 0;
104 }
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原文地址:https://www.cnblogs.com/xxx0624/p/3215600.html