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POJ3335 POJ3130 [半平面交]

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");    }}

 

POJ3335 POJ3130 [半平面交]