uva 1342 - That Nice Euler Circuit

题意：平面上有一个包含n个点的一笔画，图案是封闭的。线段可以相交，但不会部分重叠，如图，求平面被分成多少部分。

#include<iostream>#include<cstdio>#include<algorithm>#define eps 1e-8using namespace std;int sig(double a){return (a>eps)-(a<-eps);}typedef struct point{double x,y;point(double xx=0,double yy=0):x(xx),y(yy){}}vector;bool operator < (point a,point b){return a.x<b.x || (a.x==b.x && a.y<b.y);}bool operator != (point a,point b){return a.x!=b.x || a.y!=b.y;}vector operator - (point a,point b){return vector(a.x-b.x,a.y-b.y);}point operator + (point a,vector b){return point(a.x+b.x,a.y+b.y);}vector operator * (point a,double b){return vector(a.x*b,a.y*b);}double dot(vector a,vector b){return a.x*b.x+a.y*b.y;}double cross(vector a,vector b){return a.x*b.y-a.y*b.x;}bool inter(point a1,vector b1,point a2,vector b2,point &p){double a,b,t;vector v=b1-a1,w=b2-a2;if(sig(cross(v,w))==0) return 0;a=cross(v,a2-a1)*cross(v,b2-a1);b=cross(w,a1-a2)*cross(w,b1-a2);if(sig(a)>0 || sig(b)>0) return 0;vector u=a1-a2;t=cross(w,u)/cross(v,w);p=a1+v*t;return 1;}bool onseg(point a,point p,point q){if(sig(cross(p-a,p-q))==0 && sig(dot(a-p,a-q))<0) return 1;return 0;}int main(){int i,j,k,m,n,c=0;point p[305],q[50000];while(scanf("%d",&n)!=EOF && n){for(i=0;i<n;i++) {scanf("%lf%lf",&p[i].x,&p[i].y);q[i]=p[i];}m=k=0;for(i=0;i<n-3;i++){for(j=i+2;j<n-1;j++) if(inter(p[i],p[i+1],p[j],p[j+1],q[n+k])) k++;}sort(q,q+n+k);for(i=1;i<n+k;i++) if(q[m]!=q[i]) q[++m]=q[i];m++;k=n-1;for(i=0;i<m;i++){for(j=0;j<n-1;j++) if(onseg(q[i],p[j],p[j+1])) k++;}printf("Case %d: There are %d pieces.\n",++c,k-m+2);}return 0;}