Problem: Example

Input

4 4
1 2 3
1 3 3
2 3 3
2 4 3


Output

1 3

Solution (in CPP):

Solution:
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <climits>
#include <ctime>
#include <cassert>
using namespace std;

#define rep(i,a,n) for(int (i)=(a); (i)<(n); (i)++)
#define repq(i,a,n) for(int (i)=(a); (i)<=(n); (i)++)
#define repr(i,a,n) for(int (i)=(a); (i)>=(n); (i)--)
#define all(v) begin(v), end(v)
#define pb(a) push_back(a)
#define fr first
#define sc second
#define INF 2000000000
#define int long long int

#define X real()
#define Y imag()
#define EPS (1e-10)
#define EQ(a,b) (abs((a) - (b)) < EPS)
#define EQV(a,b) ( EQ((a).X, (b).X) && EQ((a).Y, (b).Y) )
#define LE(n, m) ((n) < (m) + EPS)
#define LEQ(n, m) ((n) <= (m) + EPS)
#define GE(n, m) ((n) + EPS > (m))
#define GEQ(n, m) ((n) + EPS >= (m))

typedef vector<int> VI;
typedef vector<VI> MAT;
typedef pair<int, int> pii;
typedef long long ll;

typedef complex<double> P;
typedef pair<P, P> L;
typedef pair<P, double> C;

int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0};
int const MOD = 1000000007;

namespace std {
    bool operator<(const P& a, const P& b) {
        return a.X != b.X ? a.X < b.X : a.Y < b.Y;
    }
}

// ?§?????????¨????????¨???????????????????¨???????????§???????
template <typename T>
struct Edge {
    int from, to;
    T cost;
    Edge(int s, T d) : to(s), cost(d) {}
    Edge(int f, int s, T d) : from(f), to(s), cost(d) {}

    bool operator<(const Edge &e) const {
        return cost < e.cost;
    }
    bool operator>(const Edge &e) const {
        return cost > e.cost;
    }
};
#define Graph(type) vector< vector< Edge<type> > >

template <typename T>
pair<int, vector< Edge<T> > > prim(vector< vector< Edge<T> > > &G) {
    int V = G.size(); T ans = 0;
    vector< Edge<T> > ans_e;
    vector<bool> used(V, false);
    priority_queue< Edge<T>, vector< Edge<T> >, greater< Edge<T> > > q;
    q.push(Edge<T>(-1, 0, 0));
    while(!q.empty()) {
        Edge<T> temp = q.top(); q.pop();
        if(used[temp.to]) continue;
        used[temp.to] = true;
        ans += temp.cost; if(temp.from >= 0) ans_e.pb(temp);
        rep(i,0,G[temp.to].size()) q.push(G[temp.to][i]);
    }
    rep(i,0,V) if(!used[i]) ans = numeric_limits<T>::max();
    return pair<int, vector< Edge<T> > >(ans, ans_e);
}

// Union-Find ??¨ (Verified: AtCoder Typical Contest 001 B)
struct UnionFind {
private:
    const int n;
    int __size;
    vector<int> uf;
public:
    // ????????? UnionFind uni(n) ??????????????£?¨?????????°??????
    UnionFind(int _n) : n(_n), __size(_n), uf(_n, -1) {}
    // find (??¨??????????±???????)
    int find(int x) {return (uf[x] < 0) ? x : uf[x] = find(uf[x]);}
    // x ??¨ y ???????????????????±???????????????????
    bool same(int x, int y) {return find(x) == find(y);}
    // x ????±????????????????????´???°
    int size(int x) {return -uf[find(x)];}
    // ???????????????????????????
    int size()      {return __size;}
    // x ??¨ y ????±??????????????????????
    void unite(int x, int y) {
        x = find(x); y = find(y);
        if(x == y) return;
        __size--;
        if(uf[y] < uf[x]) swap(x, y);
        uf[x] += uf[y]; uf[y] = x;
    }
    // ???????????°??¨ (???????????????)
    void debug() {for(auto e : uf) printf("UnionFind_debug: %lld\n", e);}
};

signed main() {
    int n, m; cin >> n >> m;
    Graph(int) G(n);
    vector< Edge<int> > es;
    int s, d, c;
    rep(i,0,m) {
        cin >> s >> d >> c; s--; d--;
        G[s].pb(Edge<int>(s, d, c));
        G[d].pb(Edge<int>(d, s, c));
        es.pb(Edge<int>(d, s, c));
        es.pb(Edge<int>(s, d, c));
    }
    pii ans = pii(0, 0);
    pair< int, vector< Edge<int> > > v = prim(G);
    sort(es.begin(), es.end());
    int E = es.size();
    rep(i,0,v.sc.size()) {
        UnionFind uf(n);
        int res = 0;
        for(int j=0; j<E; j++) {
            Edge<int> e = es[j];
            if(e.to == v.sc[i].from && e.from == v.sc[i].to) continue;
            if(e.to == v.sc[i].to && e.from == v.sc[i].from) continue;
            if(!uf.same(e.from, e.to)) {
                uf.unite(e.from, e.to);
                res += e.cost;
            }
        }

        int par = -1;
        for(int j=0; j<n; j++) {
            if(par == -1) par = uf.find(j);
            else if(par != uf.find(j)) res = INT_MAX;
        }

        if(res > v.fr) {
            ans.fr++;
            ans.sc += v.sc[i].cost;
        }
    }
    printf("%lld %lld\n", ans.fr, ans.sc);
    return 0;
}