Cod sursa(job #1410399)

Utilizator mirceadinoMircea Popoveniuc mirceadino Data 31 martie 2015 01:03:20
Problema Arbore partial de cost minim Scor 100
Compilator cpp Status done
Runda Arhiva educationala Marime 1.92 kb
#include<algorithm>
#include<bitset>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<deque>
#include<fstream>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<unordered_map>
#include<unordered_set>
#include<utility>
#include<vector>

using namespace std;

#ifdef HOME
const string inputFile = "input.txt";
const string outputFile = "output.txt";
#else
const string problemName = "apm";
const string inputFile = problemName + ".in";
const string outputFile = problemName + ".out";
#endif

typedef pair<int, int> PII;
const int NMAX = 200000 + 5;
const int MMAX = 400000 + 5;
const int INF = (1 << 30) - 1;

int N, M, cost;
vector<PII> V[NMAX];
int D[NMAX];
PII E[NMAX];
int dad[NMAX];
bitset<MMAX> viz;
priority_queue<PII, vector<PII>, greater<PII> > PQ;

void prim() {
    int i, x, y, z;

    for(i = 1; i <= N; i++)
        D[i] = INF;

    D[1] = 0;
    PQ.push(make_pair(0, 1));

    for(i = 1; i <= N;) {
        x = PQ.top().second;
        PQ.pop();

        if(viz[x])
            continue;

        E[i++] = make_pair(x, dad[x]);
        cost += D[x];

        viz[x] = 1;
        D[x] = 0;

        for(auto nod : V[x]) {
            y = nod.first;
            z = nod.second;

            if(!viz[y] && D[y] > z) {
                D[y] = z;
                PQ.push(make_pair(D[y], y));
                dad[y] = x;
            }
        }
    }
}

int main() {
    int i, j, x, y, z;

#ifndef ONLINE_JUDGE
    freopen(inputFile.c_str(), "r", stdin);
    freopen(outputFile.c_str(), "w", stdout);
#endif

    scanf("%d%d", &N, &M);

    for(i = 1; i <= M; i++) {
        scanf("%d%d%d", &x, &y, &z);
        V[x].push_back(make_pair(y, z));
        V[y].push_back(make_pair(x, z));
    }

    prim();

    printf("%d\n", cost);

    printf("%d\n", N - 1);

    for(i = 2; i <= N; i++)
        printf("%d %d\n", E[i].first, E[i].second);

    return 0;
}