poj 3525 二分+半平面交

//poj 3525 二分+半平面交 #include<iostream>#include<algorithm>#include<utility>#include<cstdio>#include<cstring>#include<cstdlib>#include<sstream>#include<cmath>#include<ctime>#include<vector>#include<string>#include<map>#include<set>#include<queue>#include<stack>#include<numeric>#include<limits>using namespace std;const double eps= 1e-10;const int maxn = 1024;const int inf =1012345678;struct POINT{double x,y;void read(){scanf("%lf%lf", &x,&y);}POINT(){}POINT(double a,double b){x=a; y=b;}};bool operator ==(const POINT &a,const POINT &b){if(a.x==b.x && b.y==a.y)return true;else false;}int dblcmp(double t){if(fabs(t)<eps)return 0;return t<0?-1:1;}double cross(POINT p2,POINT p3,POINT p1)//L12*L13{ //大于0则向量p1p2（p2-p1)在向量p1p3(p3-p1)顺时针方向 //小于0则向量p1p2（p2-p1） 在向量p1p3(p3-p1)逆时针方向 //等于0则在同一直线上 return (p2.x-p1.x)*(p3.y-p1.y)-(p3.x-p1.x)*(p2.y-p1.y);}inline void tuijin(POINT &p1,POINT &p2,double d)//逆时针的两个点想凸包内推进，顺时针向外//向向量的左手方向推进{// printf("d=%lf %lf %lf %lf %lf/n",d,p1.x,p1.y,p2.x,p2.y); double px,py,l; px=-p2.y+p1.y; py=p2.x-p1.x; l=sqrt(px*px+py*py); px=d*px/l; py=d*py/l; p1.x+=px; p2.x+=px; p1.y+=py; p2.y+=py; // printf("d=%lf %lf %lf %lf %lf/n",d,p1.x,p1.y,p2.x,p2.y);}inline int cutpolygon(POINT polygon[],int vcount,POINT p1,POINT p2)//凸包（多边形不行）必须逆时针方向,直线p1p2也是按照逆时针(左手)的，也就是和多边形的方向是相同的//返回被切后多边形的点数(如果只剩一条线返回的点数也可能是正的){ POINT pp[maxn]; double now[maxn]; int a=0,b=0; int i,j;int nn; for (i=0; i<vcount; i++) { now[i]=cross(p2,polygon[i],p1); if (now[i]>0) a++;//左边 要保留的点 if (now[i]<0) b++;//右边 要删除的点 } if (a==0 && b==vcount) return 0; else if (a==0 && b!=vcount) //处理只剩下一条线的情况 { nn=0; for (i=0; i<vcount; i++) if (now[i]==0)//now[i]==0; polygon[nn++]=polygon[i]; polygon[nn]=polygon[0]; return nn; } if (b==0) return vcount; else if (b!=0)//处理切割的情况 { nn=2; polygon[vcount]=polygon[0]; now[vcount]=now[0]; i=0; for (;;) { if (now[i]>0 && now[i+1]<=0) break; i=(i+1)%vcount; } //定比分点法求坐标 pp[0].x=(now[i]*polygon[i+1].x-now[i+1]*polygon[i].x)/(now[i]-now[i+1]); pp[0].y=(now[i]*polygon[i+1].y-now[i+1]*polygon[i].y)/(now[i]-now[i+1]); j=(i+1)%vcount; for (;;) { if (now[j+1]>0) break; j=(j+1)%vcount; } pp[1].x=(now[j]*polygon[j+1].x-now[j+1]*polygon[j].x)/(now[j]-now[j+1]); pp[1].y=(now[j]*polygon[j+1].y-now[j+1]*polygon[j].y)/(now[j]-now[j+1]); j=(j+1)%vcount; for(;;j=(j+1)%vcount) { pp[nn++]=polygon[j]; if (i==j) break; } for (i=0; i<nn; i++) polygon[i]=pp[i]; polygon[nn]=polygon[0]; return nn; } }int n,m,nn;POINT p[maxn],pp[maxn];double minx,miny,maxx,maxy; int main(){// freopen("in.txt","r",stdin);int i,j,k;while(scanf("%d", &n),n){double r=0,l=0; minx=inf; miny=inf; maxx=-inf; maxy=-inf;for(i=0; i<n; i++){p[i].read(); minx=min(p[i].x,minx); miny=min(p[i].y,miny); maxx=max(p[i].x,maxx); maxy=max(p[i].y,maxy);}r=max(maxx-minx,maxy-miny)/2;p[n]=p[0];double ans=0,mid;while(dblcmp(r-l)!=0){mid=(l+r)/2;for(i=0; i<=n; i++)pp[i]=p[i];nn=n;for(i=0; i<n; i++){POINT a=p[i],b=p[i+1];tuijin(a,b,mid);nn=cutpolygon(pp,nn,a,b);if(nn<=0) break;}if(nn>0){ans=mid;l=mid;}else r=mid;}printf("%lf/n",ans);}} 2011-05-02