POJ 3449 判断多边形相交

题意：给你多边形 让你判断每个多边形分别有几个多边形与它相交。

这题的输入输出恶心，另外给出正方形对角点求出另两个点的公式

已知一正方形两对角顶点A、C的坐标分别是（ax, ay）、(cx, xy)；

求B、D坐标（bx, by）、(dx, dy)；

          bx = (cx+ax+cy-ay)/2;
          by = (cy+ay+ax-cx)/2;
          dx = (cx+ax+ay-cy)/2;
          dy = (cy+ay+cx-ax)/2;

再就没什么主意的地方，按照模拟吧。我的代码4768B，应该是现在写过最长的吧，这代码莫名其妙C++ CE了要G++。

#include 
#include
#include
#include
using namespace std;
typedef double PointType;
struct point
{
    PointType x,y;
};
struct polygon
{
    int num,ansnum;
    point p[22];
    char ans[30],name;
};
int n;
polygon data[30];
PointType Direction(point pi,point pj,point pk) //判断向量PiPj在向量PiPk的顺逆时针方向 +顺-逆0共线
{
    return (pj.x-pi.x)*(pk.y-pi.y)-(pk.x-pi.x)*(pj.y-pi.y);
}
bool On_Segment(point pi,point pj,point pk)
{
    if(pk.x>=min(pi.x,pj.x)&&pk.x<=max(pi.x,pj.x)&&pk.y>=min(pi.y,pj.y)&&pk.y<=max(pi.y,pj.y))
        return 1;
    return 0;
}
bool Segment_Intersect(point p1,point p2,point p3,point p4)
{
    PointType d1=Direction(p3,p4,p1),d2=Direction(p3,p4,p2),d3=Direction(p1,p2,p3),d4=Direction(p1,p2,p4);
    if(((d1>0&&d2<0)||(d1<0&&d2>0))&&((d3>0&&d4<0)||(d3<0&&d4>0)))
        return 1;
    if(d1==0&&On_Segment(p3,p4,p1))
        return 1;
    if(d2==0&&On_Segment(p3,p4,p2))
        return 1;
    if(d3==0&&On_Segment(p1,p2,p3))
        return 1;
    if(d4==0&&On_Segment(p1,p2,p4))
        return 1;
    return 0;
}
void getcoordinate(string a,double &x,double &y)
{
    int b[2]= {0,0};
    int f=0;
    for(int i=0,j=0; i='0'&&a[i]<='9')
        {
            if(f)
                b[j]=b[j]*10-(a[i]-'0');
            else
                b[j]=b[j]*10+a[i]-'0';
        }
        else if(a[i]==',')
            j++,f=0;
    }
    x=b[0],y=b[1];
}
void getp(int num)
{
    double x,y;
    string coordinate;
    for(int i=0; i>coordinate;
        getcoordinate(coordinate,x,y);
        data[n].p[i].x=x,data[n].p[i].y=y;
    }
}
void getnum(string s)
{
    int x,y;
    string coordinate;
    if(s=="line")
        data[n].num=2,getp(data[n].num);
    if(s=="triangle")
        data[n].num=3,getp(data[n].num);
    if(s=="square")
    {
        double x1[4],y1[4];
        data[n].num=4;
        cin>>coordinate;
        getcoordinate(coordinate,x1[0],y1[0]);
        cin>>coordinate;
        getcoordinate(coordinate,x1[2],y1[2]);
        x1[1]=(x1[2]+x1[0]+y1[2]-y1[0])/2;
        y1[1]=(y1[2]+y1[0]+x1[0]-x1[2])/2;
        x1[3]=(x1[2]+x1[0]-y1[2]+y1[0])/2;
        y1[3]=(y1[2]+y1[0]-x1[0]+x1[2])/2;
        for(int i=0; i<4; i++)
            data[n].p[i].x=x1[i],data[n].p[i].y=y1[i];
    }
    if(s=="rectangle")
    {
        double x1[4],y1[4],x2,y2;
        data[n].num=4;
        cin>>coordinate;
        getcoordinate(coordinate,x1[0],y1[0]);
        cin>>coordinate;
        getcoordinate(coordinate,x1[1],y1[1]);
        cin>>coordinate;
        getcoordinate(coordinate,x1[2],y1[2]);
        x2=x1[2]-x1[1],y2=y1[2]-y1[1];
        x1[3]=x1[0]+x2,y1[3]=y1[0]+y2;
        for(int i=0; i<4; i++)
            data[n].p[i].x=x1[i],data[n].p[i].y=y1[i];
    }
    if(s=="polygon")
    {
        int t;
        cin>>t;
        data[n].num=t,getp(t);
    }
}
bool judge(polygon a,polygon b)
{
    for(int i=0; i>c)
        {
            if(c=='.')
                return 0;
            if(c=='-')
                break;
            cin>>s;
            data[n].name=c;
            getnum(s);
            n++;
        }
        solve();
        output();
    }
    return 0;
}