【半平面交】[POJ2451]Uyuw's Concert

来源：互联网发布：巫妖王之怒3.35数据库编辑：程序博客网时间：2024/04/20 11:56
题目大意

在一个10000*10000的区域内，求半平面交的面积。
分析

半平面交模板题。
#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;#define MAXN 20000#define EPS 1e-8int n,m;void Read(int &x){    char c;    while(c=getchar(),c!=EOF)        if(c>='0'&&c<='9'){            x=c-'0';            while(c=getchar(),c>='0'&&c<='9')                x=x*10+c-'0';            ungetc(c,stdin);            return;        }}struct point{    double x,y;    point(double x=0,double y=0):x(x),y(y){    }    point operator+(const point &a)const{        return point(x+a.x,y+a.y);    }    point operator-(const point &a)const{        return point(x-a.x,y-a.y);    }    point operator*(const point &a)const{        return point(x*a.x-y*a.y,x*a.y+y*a.x);    }    void operator-=(const point&b){        *this=*this-b;    }}poly[MAXN+10];struct Line{    point p,v;    double ang;    Line(){    }    Line(const point &p,const point &v):p(p),v(v){        ang=atan2(v.y,v.x);    }    bool operator<(const Line &b)const{        return ang<b.ang;    }}L[MAXN+10];double cross(point a,point b){    return a.x*b.y-a.y*b.x;}point Get_intersection(const Line &a,const Line &b){    point p=a.p-b.p;    double t=cross(b.v,p)/cross(a.v,b.v);    return a.p+a.v*t;}bool Onleft(const Line &a,const point &b){    return cross(a.v,b-a.p)>EPS;}Line q[MAXN+10];point p[MAXN+10];int halfplaneintersection(Line *L,point *poly){    int fr,bk,i;    sort(L+1,L+n+1);    q[fr=bk=0]=L[1];    for(i=2;i<=n;i++){        while(fr<bk&&!Onleft(L[i],p[bk-1]))            bk--;        while(fr<bk&&!Onleft(L[i],p[fr]))            fr++;        q[++bk]=L[i];        if(fabs(cross(q[bk-1].v,q[bk].v))<EPS){            bk--;            if(Onleft(q[bk],L[i].p))                q[bk]=L[i];        }        if(fr<bk)            p[bk-1]=Get_intersection(q[bk],q[bk-1]);    }    while(fr<bk&&!Onleft(q[fr],p[bk-1]))        bk--;    if(bk-fr<=1)        return 0;    p[bk]=Get_intersection(q[fr],q[bk]);    for(i=fr;i<=bk;i++)        poly[++m]=p[i];    return m;}void read(){    Read(n);    double x1,x2,y1,y2;    for(int i=1;i<=n;i++){        scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);        L[i]=Line(point(x1,y1),point(x2-x1,y2-y1));    }    L[++n]=Line(point(0,10000),point(-10000,0));    L[++n]=Line(point(0,0),point(10000,0));    L[++n]=Line(point(0,0),point(0,-10000));    L[++n]=Line(point(10000,0),point(0,10000));}double ans;void print(){    int i;    poly[m+1]=poly[1];    for(i=1;i<=m;i++)        ans+=cross(poly[i],poly[i+1])/2;    printf("%.1lf",ans);}int main(){    read();    halfplaneintersection(L,poly);    print();}
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