凸包，多边形面积，线段在多边形内的判定。

zoj3570  Lott's Seal http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4569

凸包，多边形面积，线段在多边形内的判定。

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
const double eps=1e-8;
const int M=100010;
struct point{
    double x,y;
}p[M],q[M],res[M];
bool operator < (const point &l, const point &r) {
    return l.y < r.y || (fabs(l.y- r.y)<eps && l.x < r.x);
}
class ConvexHull { //凸包
    bool mult(point sp, point ep, point op) {//>包括凸包边上的点,>=不包括
        return (sp.x - op.x) * (ep.y - op.y)>= (ep.x - op.x) * (sp.y - op.y);
    }
public:
    int graham(int n,point p[],point res[]) {//多边形点个数和点数组，凸包存res
        sort(p, p + n);
        if (n == 0) return 0;
        res[0] = p[0];
        if (n == 1) return 1;
        res[1] = p[1];
        if (n == 2) return 2;
        res[2] = p[2];
        int top=1;
        for (int i = 2; i < n; i++) {
            while (top && mult(p[i], res[top], res[top-1])) top--;
            res[++top] = p[i];
        }
        int len = top;
        res[++top] = p[n - 2];
        for (int i = n - 3; i >= 0; i--) {
            while (top!=len && mult(p[i], res[top],res[top-1])) top--;
            res[++top] = p[i];
        }
        return top; // 返回凸包中点的个数
    }
} tubao;
class PolygonJudge { //任意多边形判定
#define zero(x) (((x)>0?(x):-(x))<eps)
    double xmult(point p1,point p2,point p0) {
        return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
    }
    inline int opposite_side(point p1,point p2,point l1,point l2) {
        return xmult(l1,p1,l2)*xmult(l1,p2,l2)<-eps;
    }
    inline int dot_online_in(point p,point l1,point l2) {
        return zero(xmult(p,l1,l2))&&(l1.x-p.x)*(l2.x-p.x)<eps&&(l1.y-p.y)*(l2.y-p.y)<eps;
    }
public:
    int inside_polygon(point q,int n,point p[],int on_edge=1) {//判点在任意多边形内,顶点按顺时针或逆时针给出
        point q2;
        const int offset=120000;//on_edge表示点在多边形边上时的返回值,offset为多边形坐标上限
        int i=0,cnt;
        while (i<n)
            for (cnt=i=0,q2.x=rand()+offset,q2.y=rand()+offset; i<n; i++)
                if(zero(xmult(q,p[i],p[(i+1)%n]))&&(p[i].x-q.x)*(p[(i+1)%n].x-q.x)<eps&&(p[i].y-q.y)*(p[(i+1)%n].y-q.y)<eps) return on_edge;
                else if (zero(xmult(q,q2,p[i]))) break;
                else if (xmult(q,p[i],q2)*xmult(q,p[(i+1)%n],q2)<-eps&&xmult(p[i],q,p[(i+1)%n])*xmult(p[i],q2,p[(i+1)%n])<-eps) cnt++;
        return cnt&1;
    }
    //判线段在任意多边形内,顶点按顺时针或逆时针给出,与边界相交返回1（有一个定点或两个定点在边界上，用此函时需用点在多边形内的函数）
    int inside_polygon(point l1,point l2,int n,point* p) {
        point t[M],tt;
        int i,j,k=0;
        if (!inside_polygon(l1,n,p)||!inside_polygon(l2,n,p)) return 0;
        for (i=0; i<n; i++)
            if (opposite_side(l1,l2,p[i],p[(i+1)%n])&&opposite_side(p[i],p[(i+1)%n],l1,l2)) return 0;
            else if (dot_online_in(l1,p[i],p[(i+1)%n])) t[k++]=l1;
            else if (dot_online_in(l2,p[i],p[(i+1)%n])) t[k++]=l2;
            else if (dot_online_in(p[i],l1,l2)) t[k++]=p[i];
        for (i=0; i<k; i++)
            for (j=i+1; j<k; j++) {
                tt.x=(t[i].x+t[j].x)/2;
                tt.y=(t[i].y+t[j].y)/2;
                if (!inside_polygon(tt,n,p)) return 0;
            }
        return 1;
    }
} gx;
class Area { //面积
    double xmult(point p1,point p2,point p0) {//计算cross product (P1-P0)x(P2-P0)
        return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
    }
    double xmult(double x1,double y1,double x2,double y2,double x0,double y0) {
        return (x1-x0)*(y2-y0)-(x2-x0)*(y1-y0);
    }
public:
    double area_triangle(point p1,point p2,point p3) {//计算三角形面积,输入三顶点
        return fabs(xmult(p1,p2,p3))/2;
    }
    double area_triangle(double x1,double y1,double x2,double y2,double x3,double y3) {
        return fabs(xmult(x1,y1,x2,y2,x3,y3))/2;
    }
    double area_triangle(double a,double b,double c) {//计算三角形面积,输入三边长
        double s=(a+b+c)/2;
        return sqrt(s*(s-a)*(s-b)*(s-c));
    }
    double area_polygon(int n,point p[]) {//计算多边形面积,顶点按顺时针或逆时针给出
        double s1=0,s2=0;
        for (int i=0; i<n; i++){
            s1+=p[(i+1)%n].y*p[i].x;
            s2+=p[(i+1)%n].y*p[(i+2)%n].x;
        }
        return fabs(s1-s2)/2;
    }
} area;
int main(){
    double X,Y,R,S;
    int n=12,m;
    while(~scanf("%lf%lf",&X,&Y)){
        scanf("%d",&m);
        for(int i=0;i<m;i++){
            scanf("%lf%lf",&q[i].x,&q[i].y);
        }
        scanf("%lf%lf",&R,&S);
        for(int i=0;i<6;i++){
            p[i].x=X;
            p[i].y=Y;
        }
        double tmp=R*sqrt(3.0)/2;
        p[0].x+=R;
        p[0].y+=0;
        p[1].x+=R+R/2;
        p[1].y+=tmp;
        p[2].x+=R/2;
        p[2].y+=tmp;
        p[3].x+=0;
        p[3].y+=tmp*2;
        p[4].x+=-R/2;
        p[4].y+=tmp;
        p[5].x+=-R-R/2;
        p[5].y+=tmp;
        for(int i=6;i<n;i++){
            p[i].x=2*X-p[i-6].x;
            p[i].y=2*Y-p[i-6].y;
        }
        m=tubao.graham(m,q,res);
        res[m]=res[0];
        bool flag=true;
        for(int i=0;i<m;i++){
            if(!gx.inside_polygon(res[i],res[i+1],n,p)){
                flag=false;
                break;
            }
        }
        if(flag){
            double six=area.area_polygon(n,p);
            double in=area.area_polygon(m,res);
            if(six-in>S+eps){
                puts("Succeeded.");
                continue;
            }
        }
        puts("Failed.");
    }
    return 0;
}

end