Example of How to Use SHORTEST_PATH_T()
The following program shows how the function SHORTEST_PATH_T()
can be used to compute single source shortest paths in a directed graph.
Remark: The graph algorithms in LEDA are generic, that is, they
accept graphs as well as parameterized
graphs.
In order to use the template function SHORTEST_PATH_T()
we need to include <LEDA/templates/shortest_path.t>. We
use the LEDA number type integer
as the number type NT for
the edge costs. In constrast to the C++ number type int,
LEDA's integer does not overflow.
#include <LEDA/graph/graph.h>
#include <LEDA/graph/templates/shortest_path.h>
#include <LEDA/numbers/integer.h>
using namespace leda;
In main() we first create a simple graph G
with four nodes and five edges. We use an edge_array<integer>
cost to store the costs of the edges of G.
int main()
{
graph G;
node n0=G.new_node(); node n1=G.new_node();
node n2=G.new_node(); node n3=G.new_node();
edge e0=G.new_edge(n0,n1); edge e1=G.new_edge(n0,n3);
edge e2=G.new_edge(n1,n2); edge e3=G.new_edge(n2,n3);
edge e4=G.new_edge(n3,n1);
edge_array<integer> cost(G);
cost[e0]=1; cost[e1]=-1; cost[e2]=-1;
cost[e3]=2; cost[e4]=1;
The node_array<edge>
pred and the node_array<integer> dist are used
for the result of SHORTEST_PATH_T(). If there are no negative
cost cycles reachable from n0, pred[v] will contain
the last edge on a shortest path from the source node n0 to
v. This allows a construction of the complete shortest path.
dist[v] will contain the length of a shortest path from n0
to v. If there are negative cost cycles, SHORTEST_PATH_T()
returns false. In this case all dist-values
are unspecified. One can use the functions for checking
the results of an SSSP algorithm to get more details about the result.
node_array<edge> pred(G);
node_array<integer> dist(G);
bool no_negative_cycle_reachable_from_s = SHORTEST_PATH_T(G,n0,cost,dist,pred);
if (no_negative_cycle_reachable_from_s) {
node v;
forall_nodes(v,G) {
G.print_node(v);
if (pred[v]==nil&&v!=n0)
std::cout << " is not reachable from s!" << std::endl;
else if (v!=n0) {
G.print_edge(pred[v]);
std::cout << " " << dist[v] << std::endl;
}
else std::cout << " was source node" << std::endl;
}
}
else std::cout << "Negative cycle reachable from s!"
<< " Result invalid!" << std::endl;
return 0;
}
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