POJ 1410 判断线段相交点在多边形内外

题意：判断一线段与矩形是否相交。需要注意的是输入可能不是按照左上右下的顺序，如果线段两个端点都在举行内的话也算相交。

这题分为判断线段与4边是否有交点，如果没有判断两点是否在矩形内就可以了。我用的方法是射线法。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef double PointType;
struct point
{
    PointType x,y;
};
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;
}
int Pandingdian(point a,int n,point *polygon)//1在多边形上 2在多边形外 0在多边形内
{
    point b;
    b.x=-9999999;
    b.y=a.y;
    int sum=0;polygon[n]=polygon[0];
    for(int i=1; i<=n; i++)
        if(polygon[i].y-polygon[i-1].y!=0&&Segment_Intersect(a,b,polygon[i],polygon[i-1]))
        {
            if(Direction(a,polygon[i],polygon[i-1])==0)
                return 1;
            sum++;
        }
    if(sum&1)
        return 0;
    return 2;
}
int main()
{
    point a,b,lt,rb,lb,rt,polygon[10];
    int n;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,<.x,<.y,&rb.x,&rb.y);
        if(lt.x>rb.x)
            swap(lt.x,rb.x);
        if(lt.y<rb.y)
            swap(lt.y,rb.y);
        rt.x=rb.x,rt.y=lt.y;
        lb.x=lt.x,lb.y=rb.y;
        polygon[0]=lb,polygon[1]=rb,polygon[2]=rt,polygon[3]=lt,polygon[4]=lb;
        int ans=0;
        for(int i=1;i<5;i++)
        if(Segment_Intersect(a,b,polygon[i],polygon[i-1]))
        {
            puts("T"),ans=1;
            break;
        }
        if(ans)
        continue;
        if(Pandingdian(a,4,polygon)==0&&Pandingdian(b,4,polygon)==0)
        {
            puts("T");
            continue;
        }
        puts("F");
    }
    return 0;
}