[fzu 2273]判断两个三角形的位置关系

首先判断是否相交，就是枚举3*3对边的相交关系。

如果不相交，判断包含还是相离，就是判断点在三角形内还是三角形外。两边各判断一次。

//http://acm.fzu.edu.cn/problem.php?pid=2273
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;

const double eps=1e-;
const double pi=acos(-1.0);
int sgn(double x)
{
    if (fabs(x)<eps) return ;
    if (x<) return -;
    return ;
}
struct Point
{
    double x,y;
    Point() {}
    Point(double _x,double _y)
    {
        x=_x;
        y=_y;
    }
    Point operator +(const Point &b) const
    {
        return Point(x+b.x,y+b.x);
    }
    Point operator -(const Point &b) const
    {
//        return Point(x-b.x,y-b.x);
        return Point(x-b.x,y-b.y);
    }
    double operator ^(const Point &b) const
    {
        return x*b.y-y*b.x;
    }
    double operator *(const Point &b) const
    {
        return x*b.x+y*b.y;
    }
    Point operator /(const double b) const
    {
        return Point(x/b,y/b);
    }
};

struct Line
{
    Point s,e;
    Line() {}
    Line(Point _s,Point _e)
    {
        s=_s;
        e=_e;
    }
};

double dist(Point a,Point b)
{
    return sqrt((a-b)*(a-b));
}
bool inter(Line l1,Line l2)
{
    return
        max(l1.s.x,l1.e.x)>=min(l2.s.x,l2.e.x) &&
        max(l2.s.x,l2.e.x)>=min(l1.s.x,l1.e.x) &&
        max(l1.s.y,l1.e.y)>=min(l2.s.y,l2.e.y) &&
        max(l2.s.y,l2.e.y)>=min(l1.s.y,l1.e.y) &&
        sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e))<= &&
        sgn((l1.s-l2.e)^(l2.s-l2.e))*sgn((l1.e-l2.e)^(l2.s-l2.e))<=;
}
Point o;
bool _cmp(Point p1,Point p2)
{
    double tmp=(p1-o)^(p2-o);
    if (sgn(tmp)>) return true;
    else if (sgn(tmp)== && sgn(dist(p1,o)-dist(p2,o))<=) return true;
    else return false;
}

bool OnSeg(Point P,Line L)
{
    return
        sgn((L.s-P)^(L.e-P))== &&
        sgn((P.x-L.s.x)*(P.x-L.e.x))<= &&
        sgn((P.y-L.s.y)*(P.y-L.e.y))<=;
}
int inConvexPoly(Point a,Point p[],int n)
{
    for (int i=;i<n;i++)
    {
        if (sgn((p[i]-a)^(p[(i+)%n]-a))<) return -;  // out
        else if (OnSeg(a,Line(p[i],p[(i+)%n]))) return ;  // on
    }
    return ; // in
}

Point p[];

int main()
{
    int t;
    scanf("%d",&t);
    while (t--)
    {
        for (int i=;i<;i++) scanf("%lf%lf",&p[i].x,&p[i].y);
        bool xj=false;
        for (int i=;i<;i++)
        {
            for (int j=;j<;j++)
            {
                Line l1=Line(p[i],p[(i+)%]);
                Line l2=Line(p[j+],p[(j+)%+]);
                if (inter(l1,l2))
                {
                    xj=true;
                    break;
                }
            }
            if (xj) break;
        }
        if (xj)
        {
            printf("intersect\n");
            continue;
        }
        int in1=,in2=;
        o=(p[]+p[]+p[])/3.0;
        sort(p,p+,_cmp);
        o=(p[]+p[]+p[])/3.0;
        sort(p+,p+,_cmp);
        for (int i=;i<;i++)
        {
            if (inConvexPoly(p[i],p+,)==) in1++;
            if (inConvexPoly(p[i+],p,)==) in2++;
        }
        if (in1==||in2==) printf("contain\n");
        else printf("disjoint\n");
    }
    return ;
}