FZU 2273 Triangles 第八届福建省赛 （三角形面积交 有重边算相交）

Problem Description

This is a simple problem. Given two triangles A and B, you should determine they are intersect, contain or disjoint. (Public edge or point are treated as intersect.)

 Input

First line contains an integer T (1 ≤ T ≤ 10), represents there are T test cases.

For each test case: X1 Y1 X2 Y2 X3 Y3 X4 Y4 X5 Y5 X6 Y6. All the coordinate are integer. (X1,Y1) , (X2,Y2), (X3,Y3) forms triangles A ; (X4,Y4) , (X5,Y5), (X6,Y6) forms triangles B.

-10000<=All the coordinate <=10000

 Output

For each test case, output “intersect”, “contain” or “disjoint”.

 Sample Input

2

0 0 0 1 1 0 10 10 9 9 9 10

0 0 1 1 1 0 0 0 1 1 0 1

 Sample Output

disjoint

intersect

 

观察样例可以发现，如果边有重叠部分，此题也算相交。

因此我套用多边形面积交的模板http://www.cnblogs.com/Annetree/p/6535294.html

如果有重合面积，有两种情况：包含或相交，特殊判断包含即可

如果没有重合面积，也有两种情况：相交（线）和相离，特殊判断线部分重合即可

 

 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #include<cstdlib>
 #include<queue>
 #include<map>
 #include<stack>
 #include<set>

 using namespace std;

 const int maxn=;
 const int maxisn=;
 const double eps=1e-;
 const double pi=acos(-1.0);

 int dcmp(double x){
     if(x>eps) return ;
     return x<-eps ? - : ;
 }
 inline double Sqr(double x){
     return x*x;
 }

 #define zero(x)(((x)>0?(x):-(x))<eps)
 struct Point{
     double x,y;
     Point(){x=y=;}
     Point(double x,double y):x(x),y(y){};
     friend Point operator + (const Point &a,const Point &b) {
         return Point(a.x+b.x,a.y+b.y);
     }
     friend Point operator - (const Point &a,const Point &b) {
         return Point(a.x-b.x,a.y-b.y);
     }
     friend bool operator == (const Point &a,const Point &b) {
         return dcmp(a.x-b.x)==&&dcmp(a.y-b.y)==;
     }
     friend Point operator * (const Point &a,const double &b) {
         return Point(a.x*b,a.y*b);
     }
     friend Point operator * (const double &a,const Point &b) {
         return Point(a*b.x,a*b.y);
     }
     friend Point operator / (const Point &a,const double &b) {
         return Point(a.x/b,a.y/b);
     }
     friend bool operator < (const Point &a, const Point &b) {
         return a.x < b.x || (a.x == b.x && a.y < b.y);
     }
     inline double dot(const Point &b)const{
         return x*b.x+y*b.y;
     }
     inline double cross(const Point &b,const Point &c)const{
         return (b.x-x)*(c.y-y)-(c.x-x)*(b.y-y);
     }

 };

 Point LineCross(const Point &a,const Point &b,const Point &c,const Point &d){
     double u=a.cross(b,c),v=b.cross(a,d);
     return Point((c.x*v+d.x*u)/(u+v),(c.y*v+d.y*u)/(u+v));
 }
 double PolygonArea(Point p[],int n){
      if(n<) return 0.0;
      double s=p[].y*(p[n-].x-p[].x);
      p[n]=p[];
      for(int i=;i<n;i++){
         s+=p[i].y*(p[i-].x-p[i+].x);
      }
      return fabs(s*0.5);
 }
 double CPIA(Point a[],Point b[],int na,int nb){
     Point p[maxisn],temp[maxisn];
     int i,j,tn,sflag,eflag;
     a[na]=a[],b[nb]=b[];
     memcpy(p,b,sizeof(Point)*(nb+));
     for(i=;i<na&&nb>;++i){
         sflag=dcmp(a[i].cross(a[i+],p[]));
         for(j=tn=;j<nb;++j,sflag=eflag){
             if(sflag>=) temp[tn++]=p[j];
             eflag=dcmp(a[i].cross(a[i+],p[j+]));
             if((sflag^eflag)==-)
                 temp[tn++]=LineCross(a[i],a[i+],p[j],p[j+]);
         }
         memcpy(p,temp,sizeof(Point)*tn);
         nb=tn,p[nb]=p[];
     }
     if(nb<) return 0.0;
     return PolygonArea(p,nb);
 }
 double SPIA(Point a[],Point b[],int na,int nb){
     int i,j;
     Point t1[],t2[];
     double res=0.0,if_clock_t1,if_clock_t2;
     a[na]=t1[]=a[];
     b[nb]=t2[]=b[];
     for(i=;i<na;i++){
         t1[]=a[i-],t1[]=a[i];
         if_clock_t1=dcmp(t1[].cross(t1[],t1[]));
         if(if_clock_t1<) swap(t1[],t1[]);
         for(j=;j<nb;j++){
             t2[]=b[j-],t2[]=b[j];
             if_clock_t2=dcmp(t2[].cross(t2[],t2[]));
             if(if_clock_t2<) swap(t2[],t2[]);
             res+=CPIA(t1,t2,,)*if_clock_t1*if_clock_t2;
         }
     }
     return res;
     //return PolygonArea(a,na)+PolygonArea(b,nb)-res;
 }

 Point a[],b[];
 Point aa[],bb[];

 double length(Point p1,Point p2)//求边长
 {
     double res=sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
     return res;
 }

 double area_triangle(double l1,double l2,double l3)//求三角形面积
 {
     double s=(l1+l2+l3)/2.0;
     double res=sqrt(s*(s-l1)*(s-l2)*(s-l3));
     return res;
 }

 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);
 }

 int parallel(Point u1,Point u2,Point v1,Point v2)//判断平行
 {
     return zero((u1.x-u2.x)*(v1.y-v2.y)-(v1.x-v2.x)*(u1.y-u2.y));
 }

 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;
 }
 int main()
 {
     int T;
     scanf("%d",&T);
     while(T--)
     {
         for(int i=;i<;i++) scanf("%lf %lf",&a[i].x,&a[i].y);
         for(int i=;i<;i++) scanf("%lf %lf",&b[i].x,&b[i].y);
         double area_double =fabs(SPIA(a,b,,));//重合面积
         double l1=length(a[],a[]),l2=length(a[],a[]),l3=length(a[],a[]);
         double l4=length(b[],b[]),l5=length(b[],b[]),l6=length(b[],b[]);
         if(area_double>eps) //包含或相交
         {
             if(abs(area_double-area_triangle(l1,l2,l3))<eps)
                 printf("contain\n");
             else if(abs(area_double-area_triangle(l4,l5,l6))<eps)
                 printf("contain\n");
             else
                 printf("intersect\n");
         }
         else //相交或相离
         {
             bool flag=false;
             //判断是否有边重合
             for(int i=;i<;i++)
             {
                 for(int ii=i+;ii<;ii++)
                 {
                     for(int j=;j<;j++)
                     {
                         for(int jj=j+;jj<;jj++)
                         {
                             if(parallel(a[i],a[ii],b[j],b[jj]))
                                 if(dot_online_in(a[i],b[j],b[jj]))
                                 {
                                     flag=true;
                                     break;
                                 }
                         }
                         if(flag)
                             break;
                     }
                     if(flag)
                         break;
                 }
                 if(flag)
                     break;
             }
             if(flag)
                 printf("intersect\n");
             else
                 printf("disjoint\n");

         }
     }
     return ;
 }