标签:bool break name point i++ 交点 getc 自己的 print
终于写出自己的半平面交模板了.......
加入交点的地方用了直线线段相交判定
两个题一样,只不过一个顺时针一个逆时针(给出一个多边形的两种方式啦),反正那个CutPolygon是切掉左面
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <vector> using namespace std; typedef long long ll; const int N=105; const double INF=1e5; const double eps=1e-8; inline int read(){ char c=getchar();int x=0,f=1; while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1; c=getchar();} while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘; c=getchar();} return x*f; } inline int sgn(double x){ if(abs(x)<eps) return 0; else return x<0?-1:1; } struct Vector{ double x,y; Vector(double a=0,double b=0):x(a),y(b){} bool operator <(const Vector &a)const{ return sgn(x-a.x)<0||(sgn(x-a.x)==0&&sgn(y-a.y)<0); } void print(char c){printf("%c %lf %lf\n",c,x,y);} }; typedef Vector Point; Vector operator +(Vector a,Vector b){return Vector(a.x+b.x,a.y+b.y);} Vector operator -(Vector a,Vector b){return Vector(a.x-b.x,a.y-b.y);} Vector operator *(Vector a,double b){return Vector(a.x*b,a.y*b);} Vector operator /(Vector a,double b){return Vector(a.x/b,a.y/b);} bool operator ==(Vector a,Vector b){return sgn(a.x-b.x)==0&&sgn(a.y-b.y)==0;} 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;} double Len(Vector a){return sqrt(Dot(a,a));} double DisTL(Point p,Point a,Point b){ Vector v1=b-a,v2=p-a; return abs(Cross(v1,v2)/Len(v1)); } struct Line{ Point s,t; Line(){} Line(Point a,Point b):s(a),t(b){} }; bool isLSI(Line l1,Line l2){ Vector v=l1.t-l1.s,u=l2.s-l1.s,w=l2.t-l1.s; return sgn(Cross(v,u))!=sgn(Cross(v,w)); } Point LI(Line a,Line b){ Vector v=a.s-b.s,v1=a.t-a.s,v2=b.t-b.s; double t=Cross(v2,v)/Cross(v1,v2); return a.s+v1*t; } void iniPolygon(Point p[],int &n,double inf){ n=0; p[++n]=Point(-inf,-inf); p[++n]=Point(inf,-inf); p[++n]=Point(inf,inf); p[++n]=Point(-inf,inf); } Point t[N];int tn; void CutPolygon(Point p[],int &n,Point a,Point b){//get the left of a->b memset(t,0,sizeof(t));tn=0; Point c,d,e; for(int i=1;i<=n;i++){ c=p[i],d=p[i%n+1]; if(sgn(Cross(b-a,c-a))>=0) t[++tn]=c; if(isLSI(Line(a,b),Line(c,d))){ e=LI(Line(a,b),Line(c,d));//e.print(‘e‘); t[++tn]=e; } } n=tn;for(int i=1;i<=n;i++)p[i]=t[i]; } int n,m; Point p[N],q[N]; int main(int argc, const char * argv[]) { while(true){ n=read();if(n==0) break; iniPolygon(q,m,INF); for(int i=1;i<=n;i++) p[i].x=read(),p[i].y=read(); for(int i=1;i<=n;i++) CutPolygon(q,m,p[i],p[i%n+1]);//,printf("%d\n",m); if(m) puts("1");else puts("0"); } }
标签:bool break name point i++ 交点 getc 自己的 print
原文地址:http://www.cnblogs.com/candy99/p/6358753.html