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Cod sursa(job #852927)
//Dijkstra O(nlog(n)) 100p cu STL tipul SET (arbori de cautare echilibrati )
// SET - Toate operatiile se fac in O(log(n))
#include <fstream>
#include <set>
#include <vector>
#define NMax 50100
#define inf 1000000000
using namespace std;
ifstream f("dijkstra.in");
ofstream g("dijkstra.out");
int n, m, d[NMax]; vector<int> G[NMax], C[NMax];
set< pair<int, int> > T;
int i, x, y, costul;
void solve(void)
{
int i, val, x;
for(i = 2; i <= n; i++) d[i] = inf;
T.insert( make_pair(0, 1) );
while( T.size() > 0 )
{
val = (*T.begin()).first;
x = (*T.begin()).second;
T.erase(*T.begin());
for(i = 0; i < G[x].size(); i++)
if(d[ G[x][i] ] > val + C[x][i] )
{
d[ G[x][i] ] = val + C[x][i];
T.insert(make_pair(d[G[x][i]],G[x][i]));
}
}
}
int main(void)
{
f>>n>>m;
for(i = 1; i <=m ; i++)
{
f>>x>>y>>costul;
G[x].push_back(y); C[x].push_back(costul);
}
solve();
for ( int i = 2; i <= n; i++ )
if(d[i] == inf) g<<0<<" ";else g<< d[i]<<" ";
g<<'\n';
return 0;
}
////Dijkstra 40p cu memorare graf cu liste de adiacenta O(n*n)clasic nu se incadreaza in timp
//#include <fstream>
//#define NMax 50003
//#define inf 1000000000
//using namespace std;
//ifstream f("dijkstra.in");
//ofstream g("dijkstra.out");
//struct Nod
//{
// int nod, cost;
// Nod *next;
//};
//int n, m,x,y,costul;
//Nod *Vecin[NMax];
//int d[NMax], S[NMax];
//void adauga(int x, int y, int costul)
//{
// Nod *q = new Nod;
// q->nod = y;
// q->cost = costul;
// q->next = Vecin[x];
// Vecin[x] = q;
//}
//
//void dijkstra()
//{
// for ( int i = 2; i <= n; i++ ) d[i] = inf;
// int mini, pmin = 0;
// for ( int i = 1; i <= n; i++ )
// {
// mini = inf;
// for ( int j = 1; j <= n; j++ )
// if ( d[j] < mini && !S[j] ) {mini = d[j]; pmin = j;}
//
// S[pmin] = 1;
// Nod *t = Vecin[pmin];
// while ( t )
// {
// if ( d[ t->nod ] > d[pmin] + t->cost )
// d[ t->nod ] = d[pmin] + t->cost;
// t = t->next;
// }
// }
//}
//
//int main()
//{
// f>>n>>m;
// for ( int i = 1; i <= m; i++ )
// {
// f>>x>>y>>costul;
// adauga(x, y, costul);
// }
// dijkstra();
// for ( int i = 2; i <= n; i++ )
// if(d[i] == inf) g<<0<<" ";else g<< d[i]<<" ";
// g<<'\n';
//
// return 0;
//}