TUPLE+

Answer is HERE!

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int> p_t;
const int MX=300005;
map<p_t, set<int>> d;
int n,m,u[MX],v[MX],w[MX];
int main(){
    ios::sync_with_stdio(false);
    cin>>n>>m;
    for(int i=1;i<=m;i++){
        cin>>u[i]>>v[i]>>w[i];
        d[make_pair(u[i],v[i])].insert(w[i]);}
    int ans=0;
    for(int i=1;i<=m;i++){
        set<int> *W=&d[make_pair(u[i],v[i])], *V=&d[make_pair(u[i],w[i])], *U=&d[make_pair(v[i],w[i])];
        if(U->size()>V->size())swap(U,V);
        if(V->size()>W->size())swap(V,W);
        if(U->size()>V->size())swap(U,V); // 冒泡排序，找到最小的一个集合，保证求交集复杂度
        for(auto j:*U)if(V->find(j)!=V->end()&&W->find(j)!=W->end()){
            ans++;}}
    cout<<ans<<endl;
    return 0;}

Thant you! With pleasure.