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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