POJ3525+半平面交

题意：求出给定的凸多边形的内接圆

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<algorithm>using namespace std;const int maxn = 155;const int maxm = 1005;const double Min = 0.0;const double Max = 10000000000.0;const double eps = 1e-8;const double pi = acos(-1.0);struct Point{double x,y;};struct Line{Point a,b;};Point pnt[ maxn ],res[ maxm ],tp[ maxm ];double xmult( Point op,Point sp,Point ep ){return (sp.x-op.x)*(ep.y-op.y)-(sp.y-op.y)*(ep.x-op.x);}double dist( Point a,Point b ){return sqrt( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y) );}void Get_equation( Point p1,Point p2,double &a,double &b,double &c ){a = p2.y-p1.y;b = p1.x-p2.x;c = p2.x*p1.y-p1.x*p2.y;}//直线Point Intersection( Point p1,Point p2,double a,double b,double c ){double u = fabs( a*p1.x+b*p1.y+c );double v = fabs( a*p2.x+b*p2.y+c );Point tt;tt.x = (p1.x*v+p2.x*u)/(u+v);tt.y = (p1.y*v+p2.y*u)/(u+v);return tt;}//交点double GetArea( Point p[],int n ){double sum = 0;for( int i=2;i<n;i++ ){sum += xmult( p[1],p[i],p[i+1] );}return -sum/2.0;}//面积void cut( double a,double b,double c,int &cnt ){int temp = 0;for( int i=1;i<=cnt;i++ ){if( a*res[i].x+b*res[i].y+c>-eps ){//>=0tp[ ++temp ] = res[i];}else{if( a*res[i-1].x+b*res[i-1].y+c>eps ){tp[ ++temp ] = Intersection( res[i-1],res[i],a,b,c );}if( a*res[i+1].x+b*res[i+1].y+c>eps ){tp[ ++temp ] = Intersection( res[i],res[i+1],a,b,c );}}}for( int i=1;i<=temp;i++ )res[i] = tp[i];res[ 0 ] = res[ temp ];res[ temp+1 ] = res[ 1 ];cnt = temp;}bool solve( int n,double r ){for( int i=0;i<=n+1;i++ )res[ i ] = pnt[ i ];int cnt = n;for(int i=1;i<=n;i++){          double a,b,c;          Point p1,p2,p3;          p1.y=pnt[i].x-pnt[i+1].x;p1.x=pnt[i+1].y-pnt[i].y;          double k=r/sqrt(p1.x*p1.x+p1.y*p1.y);          p1.x=k*p1.x;p1.y=p1.y*k;          p2.x=p1.x+pnt[i].x;p2.y=p1.y+pnt[i].y;          p3.x=p1.x+pnt[i+1].x;p3.y=p1.y+pnt[i+1].y;         Get_equation( p2,p3,a,b,c );          cut(a,b,c,cnt);      }  if( cnt<=1 ) return false;else return true;}int main(){int n;while( scanf("%d",&n)==1,n ){for( int i=1;i<=n;i++ )scanf("%lf%lf",&pnt[i].x,&pnt[i].y);if( GetArea( pnt,n)<eps )reverse( pnt+1,pnt+1+n );pnt[0] = pnt[n];pnt[n+1] = pnt[1];double L,R,mid,ans;L = Min;R = Max;while( R-L>=eps ){//R>Lmid = (L+R)/2.0;if( solve( n,mid )==true ){ans = mid;L = mid;}else{R = mid;}}printf("%.6lf\n",ans);}return 0;}