1 /*
2 类型:多边形相交面积模板
3 */
4
5 #include<cstdio>
6 #include<iostream>
7 #include<algorithm>
8 #include<cstring>
9 #include<cmath>
10 using namespace std;
11 #define maxn 510
12 const double eps=1E-8;
13 int sig(double d){
14 return(d>eps)-(d<-eps);
15 }
16 struct Point{
17 double x,y; Point(){}
18 Point(double x,double y):x(x),y(y){}
19 bool operator==(const Point&p)const{
20 return sig(x-p.x)==0&&sig(y-p.y)==0;
21 }
22 };
23 double cross(Point o,Point a,Point b){
24 return(a.x-o.x)*(b.y-o.y)-(b.x-o.x)*(a.y-o.y);
25 }
26 double area(Point* ps,int n){
27 ps[n]=ps[0];
28 double res=0;
29 for(int i=0;i<n;i++){
30 res+=ps[i].x*ps[i+1].y-ps[i].y*ps[i+1].x;
31 }
32 return res/2.0;
33 }
34 int lineCross(Point a,Point b,Point c,Point d,Point&p){
35 double s1,s2;
36 s1=cross(a,b,c);
37 s2=cross(a,b,d);
38 if(sig(s1)==0&&sig(s2)==0) return 2;
39 if(sig(s2-s1)==0) return 0;
40 p.x=(c.x*s2-d.x*s1)/(s2-s1);
41 p.y=(c.y*s2-d.y*s1)/(s2-s1);
42 return 1;
43 }
44 //多边形切割
45 //用直线ab切割多边形p,切割后的在向量(a,b)的左侧,并原地保存切割结果
46 //如果退化为一个点,也会返回去,此时n为1
47 void polygon_cut(Point*p,int&n,Point a,Point b){
48 static Point pp[maxn];
49 int m=0;p[n]=p[0];
50 for(int i=0;i<n;i++){
51 if(sig(cross(a,b,p[i]))>0) pp[m++]=p[i];
52 if(sig(cross(a,b,p[i]))!=sig(cross(a,b,p[i+1])))
53 lineCross(a,b,p[i],p[i+1],pp[m++]);
54 }
55 n=0;
56 for(int i=0;i<m;i++)
57 if(!i||!(pp[i]==pp[i-1]))
58 p[n++]=pp[i];
59 while(n>1&&p[n-1]==p[0])n--;
60 }
61 //---------------华丽的分隔线-----------------//
62 //返回三角形oab和三角形ocd的有向交面积,o是原点//
63 double intersectArea(Point a,Point b,Point c,Point d){
64 Point o(0,0);
65 int s1=sig(cross(o,a,b));
66 int s2=sig(cross(o,c,d));
67 if(s1==0||s2==0)return 0.0;//退化,面积为0
68 if(s1==-1) swap(a,b);
69 if(s2==-1) swap(c,d);
70 Point p[10]={o,a,b};
71 int n=3;
72 polygon_cut(p,n,o,c);
73 polygon_cut(p,n,c,d);
74 polygon_cut(p,n,d,o);
75 double res=fabs(area(p,n));
76 if(s1*s2==-1) res=-res;return res;
77 }
78 //求两多边形的交面积
79 double intersectArea(Point*ps1,int n1,Point*ps2,int n2){
80 if(area(ps1,n1)<0) reverse(ps1,ps1+n1);
81 if(area(ps2,n2)<0) reverse(ps2,ps2+n2);
82 ps1[n1]=ps1[0];
83 ps2[n2]=ps2[0];
84 double res=0;
85 for(int i=0;i<n1;i++){
86 for(int j=0;j<n2;j++){
87 res+=intersectArea(ps1[i],ps1[i+1],ps2[j],ps2[j+1]);
88 }
89 }
90 return res;//assumeresispositive!
91 }
92 //hdu-3060求两个任意简单多边形的并面积
93 Point ps1[maxn],ps2[maxn];
94 int n1=3, n2=3;
95 int main(){
96 int T; scanf("%d", &T);
97 while(T--)
98 {
99 for(int i=0;i<n1;i++)
100 scanf("%lf%lf",&ps1[i].x,&ps1[i].y);
101 for(int i=0;i<n2;i++)
102 scanf("%lf%lf",&ps2[i].x,&ps2[i].y);
103 double ans=intersectArea(ps1,n1,ps2,n2);
104 //ans=fabs(area(ps1,n1))+fabs(area(ps2,n2))-ans; //容斥,求并面积
105 double Area1=area(ps1, n1), Area2=area(ps2, n2);
106 if(ans>eps && min(Area1,Area2)-ans<eps) puts("contain") ;
107 else if(ans>eps && min(Area1,Area2)-ans>eps) puts("intersect");
108 else
109 {
110 int mp=0;
111 for(int i=0; i<n1; ++i)
112 for(int j=0; j<n2; ++j)
113 if(ps1[i].x==ps2[j].x && ps1[i].y==ps2[j].y)
114 ++mp;
115 if(mp==2) puts("intersect");
116 else puts("disjoint");
117 }
118 }
119 return 0;
120 }