首先，材料的前兩個屬性可以唯一確定一個材料，合金的前兩個樹形也可以唯一確定一個材料。

那麼材料和合金都可以被看成平面上的點\((a_i,b_i)\)或\((d_i,e_i)\)

不難發現，一些材料能表示出一種合金當且僅當這個合金（在平面上的點）在選取的材料（在平面上的點）組成的凸包內。

不難發現，選取的點凸包上的邊一定滿足：所有的合金代表的點都在這條邊的某一側，或者在這條線段上（不是直線上）

那麼我們直接求個最小環就行了。

複雜度\(O(n^3+mn^2).\)

注意特判答案為 1 或 2 的情況。

#include <bits/stdc++.h>
#define db long double
#define eps 1e-7
using namespace std;
const int N = 505,M = 505;
struct point{
	db x,y;
	point(db xx=0,db yy=0){ x = xx,y = yy; }
	bool operator < (const point w) const{
		if (fabs(x-w.x) > eps) return x < w.x;
		if (fabs(y-w.y) > eps) return y < w.x;
		return 0;
	}
}a[N],b[M];
bool operator == (point a,point b){ return fabs(a.x-b.x) < eps && fabs(a.y-b.y) < eps; }
point operator + (point a,point b){ return point(a.x+b.x,a.y+b.y); }
point operator - (point a,point b){ return point(a.x-b.x,a.y-b.y); }
inline db operator * (point a,point b){ return a.x * b.y - a.y * b.x; }
inline bool left(point a,point b,point p){ return fabs((b-a) * (p-a)) > eps && ((b-a) * (p-a)) > eps;  }
int n,m,dp[N][N],ans;
inline bool check(point p1,point p2){
	if (p1 == p2) return 0;
	for (int i = 1; i <= m; ++i) if (left(p1,p2,b[i])) return 0;
	for (int i = 1; i <= m; ++i) if (fabs((p2-p1)*(b[i]-p1)) < eps){
		if (p1 == b[i] || p2 == b[i]) continue;
		db l,r;
		l = p1.x,r = p2.x; if (l > r) swap(l,r); if (l-eps > b[i].x || b[i].x > r+eps) return 0;
		l = p1.y,r = p2.y; if (l > r) swap(l,r); if (l-eps > b[i].y || b[i].y > r+eps) return 0;
	}
	return 1;
}
int main(){
	int i,j,k;
	cin >> n >> m;
	for (i = 1; i <= n; ++i) cin >> a[i].x >> a[i].y >> a[0].x; sort(a+1,a+n+1);
	for (i = 1; i <= m; ++i) cin >> b[i].x >> b[i].y >> a[0].x; sort(b+1,b+m+1);
	for (i = 1; i <= n; ++i) a[i].x *= 1000000,a[i].y *= 1000000;
	for (i = 1; i <= m; ++i) b[i].x *= 1000000,b[i].y *= 1000000;
	n = unique(a+1,a+n+1) - (a+1);
	m = unique(b+1,b+m+1) - (b+1);
	for (i = 1; i <= n; ++i) for (j = 1; j <= n; ++j) dp[i][j] = 10000000;
	ans = 10000000;
	for (i = 1; i <= n; ++i) for (j = 1; j <= n; ++j)
		if (i != j && check(a[i],a[j])) dp[i][j] = 1;
	for (k = 1; k <= n; ++k) for (i = 1; i <= n; ++i) for (j = 1; j <= n; ++j)
		dp[i][j] = min(dp[i][j],dp[i][k] + dp[k][j]);
	for (i = 1; i <= n; ++i) ans = min(ans,dp[i][i]);
	if (n == 1 && m == 1 && a[1] == b[1]) ans = 1;
	if (ans > n) cout << -1 << '\n'; else cout << ans << '\n';
	return 0;
}