蒟蒻代码，大体思路是把三元组看做一个2000进制的数字，枚举每个三元组把他当做$(a,b,c)$，然后找所有$(b,c)$开头的就可以找到$d$。

我这里做了一个排序，然后提前处理了第一个$(b,c)$的位置，就不需要再次枚举。但是不知道怎么算时间复杂度，感觉如果特殊构造的数据应该会被hack掉呢？

求解平均时间复杂度，求大佬hack

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<unordered_map>
#define N 222005
#define int long long
using namespace std;
unordered_map <int, bool> s;
unordered_map <int, int> pos;
struct node
{
	int a;
	int b;
	int c;
	int beg;
}arr[50005];
bool cmp( node s1, node s2 )
{
	if ( s1.a == s2.a )
	{
		if ( s1.b == s2.b )
		{
			return s1.c < s2.c;
		}
		return s1.b < s2.b;
	}
	return s1.a < s2.a;
}
template <typename T>
void read ( T &x ) 
{
    x = 0;
	bool f = 0;
	char c = getchar();
    for( ; !isdigit(c); c=getchar() ) f=c=='0';
    for( ; isdigit(c); c=getchar()) x=x*10+c-'0';
    if( f ) x = -x;
}
void read( int  &x ) {
    x=0;
	bool f=0;
	char c=getchar();
    for(;!isdigit(c);c=getchar()) f=c=='0';
    for(;isdigit(c);c=getchar()) x=x*10+c-'0';
    if(f) x=-x;
}
signed main()
{
	int n = 0, m = 0;
	read( n ), read( m );
	for ( int i = 1; i <= m; i ++ )
	{
		int a = 0, b = 0, c = 0;
		read( a ), read( b ), read( c );
		arr[i].a = a, arr[i].b = b, arr[i].c = c, arr[i].beg = a * 2000 + b;
		int d = a * 4000000 + b * 2000 + c;
		s[d] = 1;
	}
	sort( arr+1, arr+1+m, cmp );
	for ( int i = 1; i <= m; i ++ )
	{
		if ( arr[i].beg != arr[i-1].beg )
		{
			pos[arr[i].beg] = i;
		}
	}
	long long ans = 0;
	for ( int i = 1; i <= m; i ++ )
	{
		int las = arr[i].b * 2000 + arr[i].c;
		for ( int j = pos[las]; ( arr[j].beg == las ) && ( j <= m ); j ++ )
		{
			if ( arr[i].c >= arr[j].c )
			{
				continue;
			}
			int sec = arr[i].a * 4000000 + arr[i].b * 2000 + arr[j].c; 
			int thd = arr[i].a * 4000000 + arr[i].c * 2000 + arr[j].c;
			if( s[sec] == 1 && s[thd] == 1 )
			{
				ans ++;
			}
		}
	}
	cout << ans << "\n";
	return 0;
}
/*

2 4 5 6
2 4 5 9

9 11
2 4 5
2 5 6
4 5 6
2 4 6
1 3 5
1 3 6
3 5 6
1 5 6
4 6 9
5 6 9
4 5 9


*/