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.)

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

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

#define _CRT_SECURE_NO_DEPRECATE

#include <iostream>

#include<vector>

#include<algorithm>

#include<cstring>

#include<bitset>

#include<set>

#include<map>

#include<cmath>

#include<queue>

using namespace std;

#define N_MAX 10000+4

#define MOD 1000000007

#define INF 0x3f3f3f3f

#define EPS 1e-8

typedef long long ll;

struct point {

    double x, y;

    point(double x = , double y = ) :x(x), y(y) {}

    point operator + (point p) { return point(x + p.x, y + p.y); }

    point operator - (point p) { return point(x - p.x, y - p.y); }

    point operator * (double a) { return point(a*x, a*y); }

    double norm() { return x*x + y*y; }

    bool operator < (const point &p) const{

        return x != p.x ? x + EPS < p.x : y + EPS < p.y;

    }

};

double dot(point a,point b) {

    return a.x*b.x + a.y*b.y;

}

double det(point a,point b) {

    return a.x*b.y - a.y*b.x;

}

typedef vector<point>polygon;

int contain( polygon g, point p) {//1代表在多边形内,0代表在多边形外,2在边上

    int n = g.size(); bool x = ;

    for (int i = ; i < n;i++) {

        point a = g[i] - p, b = g[(i + ) % n] - p;

        if (fabs(det(a, b)) < EPS&&dot(a, b) < EPS)return ;

        if (a.y > b.y)swap(a, b);

        if (a.y<EPS&&b.y>EPS&&det(a, b) > EPS)x = !x;

    }

    return x ?  : ;

}

int ccw(point a,point b,point c) {//顺时针转还是逆时针转

    point x = b - a, y = c - a;

    if(det(x, y)>EPS)return ;

    if (det(x, y) < -EPS)return -;

    if (dot(a, b) < -EPS)return ;

    if (a.norm() < b.norm())return -;

    return ;

}

int main() {

    int t; scanf("%d",&t);

    while (t--) {

        polygon T1, T2;

        T1.resize(); T2.resize();

        for (int i = ; i < ; i++)

            scanf("%lf%lf",&T1[i].x,&T1[i].y);

        for (int i = ; i < ; i++)

            scanf("%lf%lf", &T2[i].x, &T2[i].y);

        bool is_wai = , is_nei = ,flag=;

        for (int i = ; i < ;i++) {

            int c = contain(T1, T2[i]);

            if (c==) {

                is_nei = ;

            }

            else if(c==)is_wai = ;

            else { flag = ; break; }//在边上一定相交

        }

        if (flag||(is_wai&&is_nei)) { puts("intersect"); continue; }

        if (is_nei && !is_wai) { puts("contain"); }

        else {

            is_nei = ,is_wai=;

            for (int i = ; i < ;i++) {

                int c = contain(T2,T1[i]);

                if (c == )is_nei = ;

                if (c == )is_wai = ;

            }

            if (is_nei&&!is_wai)puts("contain");

            if (is_nei&&is_wai)puts("intersect");

            else {//六个点都在另外三角形的外部，判断是否有边相交的情况

                bool is_intersect = ;

                point a = T1[], b = T1[];

                for (int i = ; i < ; i++) {

                    point c = T2[i], d = T2[(i + ) % ];

                    if (ccw(a, b, c)*ccw(a, b, d) <=  && ccw(c, d, a)*ccw(c, d, b) <= ){

                        is_intersect = ; break;

                     }

                }

                if (is_intersect)puts("intersect");

                else puts("disjoint");

            }

        }

    }

    return ;

}