Cod sursa(job #2163272)

Utilizator Andrei17Andrei Pascu Andrei17 Data 12 martie 2018 17:36:00
Problema Arbore partial de cost minim Scor 90
Compilator cpp Status done
Runda Arhiva educationala Marime 1.91 kb
#include <fstream>
#include <vector>
#include <climits>

using namespace std;

ifstream in("apm.in");
ofstream out("apm.out");

const int N = 200002;

int n, m, nh;
int d[N], h[N], poz[N], pred[N];
bool viz[N];
vector <int> a[N], c[N];

/// poz[i] - al catelea in heap este elementul i din vector
/// h[i] - al catelea din vector este elementul i din heap

void citire() {
    in >> n >> m;

    int x, y, z;
    for (int i = 0; i < m; i++) {
        in >> x >> y >> z;
        a[x].push_back(y);
        a[y].push_back(x);
        c[x].push_back(z);
        c[y].push_back(z);
    }

    in.close();
}

inline void myswap(int p, int q) {
    swap(h[p], h[q]);
    poz[ h[p] ] = p;
    poz[ h[q] ] = q;
}

void urca(int p) {
    while (p > 1 && d[ h[p] ] < d[ h[p >> 1] ]) {
        myswap(p, p >> 1);
        p >>= 1;
    }
}

void coboara(int p) {
    int q1 = 2*p, q2 = 2*p + 1, best = p;

    if (q1 <= nh && d[ h[best] ] > d[ h[q1] ]) best = q1;
    if (q2 <= nh && d[ h[best] ] > d[ h[q2] ]) best = q2;

    if (best != p) {
        myswap(p, best);
        coboara(best);
    }
}

void sterge(int p) {
    myswap(p, nh--);
    urca(p);
    coboara(p);
}

void apm(int x0) {
    nh = n;
    for (int i = 1; i <= n; i++) {
        d[i] = INT_MAX;
        h[i] = poz[i] = i;
    }

    d[x0] = 0;
    urca(poz[x0]);
    while (nh > 0 && d[ h[1] ] != INT_MAX) {
        int x = h[1];
        sterge(1);
        viz[x] = true;
        for (int i = 0; i < a[x].size(); i++) {
            int y = a[x][i], cost = c[x][i];
            if (viz[y]) continue;
            if (cost < d[y]) {
                d[y] = cost;
                pred[y] = x;
                urca(poz[y]);
            }
        }
    }
}

int main()
{
    citire();
    apm(1);
    int sum = 0;
    for (int i = 2; i <= n; i++) sum += d[i];
    out << sum << '\n' << n - 1 << '\n';
    for (int i = 2; i <= n; i++) {
        out << pred[i] << ' ' << i << '\n';
    }
    out.close();
}