Cod sursa(job #1912434)

Utilizator Kln1000Ciobanu Bogdan Kln1000 Data 8 martie 2017 08:50:48
Problema Arbore partial de cost minim Scor 100
Compilator cpp Status done
Runda Arhiva educationala Marime 1.99 kb
#include <iostream>
#include <fstream>
#include <algorithm>

using namespace std;

class parser{
    public:
        parser() {}
        parser(const char *file_name){
            input_file.open(file_name,std::ios::in | std::ios::binary);
            input_file.sync_with_stdio(false);
            index&=0;
            input_file.read(buffer,SIZE);}
        inline parser &operator >>(int &n){
            for (;buffer[index]<'0' or buffer[index]>'9';inc());
            n&=0;
            sign&=0;
            sign|=(buffer[index-1]=='-');
            for (;'0'<=buffer[index] and buffer[index]<='9';inc())
                n=(n<<1)+(n<<3)+buffer[index]-'0';
            n^=((n^-n)&-sign);
            return *this;}
        ~parser(){
            input_file.close();}
    private:
        std::fstream input_file;
        static const int SIZE=0x400000; ///4MB
        char buffer[SIZE];
        int index,sign;
        inline void inc(){
            if(++index==SIZE)
                index=0,input_file.read(buffer,SIZE);}
};

parser f ("apm.in");
ofstream t ("apm.out");

struct muchie{
    int from,to,cost;
}v[200010];

int n,m,cost_total=0,multime[200010],sol[200010];

int main()
{
    int index=0;
    f>>n>>m;
    for (int x,y,c,i=0;i<m;++i){
        f>>x>>y>>c;
        v[i]={x,y,c};
    }
    sort(v,v+m,[](muchie a,muchie b){return a.cost<b.cost;});
    for (int i=0;i<n;++i)
        multime[i]=i;
    for (int aux,i=0;i<m;++i){
        int a=v[i].from;
        while(a!=multime[a])
            aux=multime[multime[a]],multime[a]=aux,a=aux;
        int b=v[i].to;
        while(b!=multime[b])
            aux=multime[multime[b]],multime[b]=aux,b=aux;
        if(a!=b){
            cost_total+=v[i].cost;
            sol[index++]=i;
            multime[a]=b;
            if(index==(n-1))   break;
        }
    }
    t<<cost_total<<'\n'<<index<<'\n';
    for (int i=0;i<index;++i)
        t<<v[sol[i]].from<<" "<<v[sol[i]].to<<'\n';
    return 0;
}