POJ3335 POJ3130 [半平面交]

Posted 2020-08-29 Candy?

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");
    }
}