double leftovers_cost(const TSPInstance *instance, const int *partialSolution, int m, tree_node *node)
{
/* Calcula o valor máximo que os vértices faltantes agregarão ao grafo <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<? */
int i = 0, j = 0;
//node *mst = NULL;
//int left_n = 0;
int *used = NULL;
bool already_in;
used = new int[instance->n];
for (i = 0; i < instance->n; i++)
{
already_in = 0;
for (j = 0; j < m; j++)
{
if (partialSolution[j] == i)
{
already_in = 1;
}
}
if (already_in == 0)
{
used[i] = 0;
}
else
{
used[i] = 1;
}
}
//left_n = instance->n - m;
return mst_prim(instance, partialSolution, 0);
//return sum_mst_pri; //asoidjaoidjs j
}
int main()
{
int n,i;
// code to take number of vertices
printf("Enter number of vertceis\n");
scanf("%d",&n);
// code to make adjacent list and display
printf("\nInstruction:\n\nfirst enter adjacent node and give one space and then enter wieght of edge \nand then do sam for next nodes and if adajcent finish \nthen enter 0 and Press Enter Key\n\n",i);
for(i=1;i<=n;i++)
{
scan_adj(i);
//printf("adj complete\n");
}
display_list(n);
//code for spanning tree
mst_prim(n);
display_edge(n);
system("pause");
return 0;
}
/* main function begins */
int main (void)
{
/* note on graph representation
-- each test input graph is treated as a unigraph, i.e. undirected
-- adjacency-matrix representation is used; thus a matrix of {x,y}
signifies a graph with x vertices, where x == y, with the presence and
weight of each connecting edge between vertices x and y denoted by the
value of cell {x,y}
-- the value of each cell is to be interpreted as follows:
-- 0: no edge/adjacency
-- i: edge/adjacency present with weight of integer i
*/
/* test unigraph 1 */
int ugraph1_size = 5; /* length/width of matrix */
int ugraph1[5][5] = /* matrix itself */
{
{ 0, 0, 1, 2, 0 },
{ 0, 0, 3, 0, 5 },
{ 2, 3, 0, 6, 4 },
{ 1, 0, 6, 0, 0 },
{ 0, 5, 4, 0, 0 }
};
/* edge for matrix, returned by prim; -1 means empty element */
struct edge ugraph1_mst[ugraph1_size*ugraph1_size];
for ( int i = 0; i < ( ugraph1_size * ugraph1_size ); i++ )
{
ugraph1_mst[i].u = -1;
ugraph1_mst[i].v = -1;
}
/* test unigraph 2 */
int ugraph2_size = 7; /* length/width of matrix */
int ugraph2[7][7] = /* matrix itself */
{
{ 0, 3, 0, 0, 0, 0, 6 },
{ 3, 0, 0, 4, 0, 0, 8 },
{ 0, 1, 0, 2, 10,0, 0 },
{ 0, 4, 2, 0, 0, 0, 5 },
{ 0, 0, 10,0, 0, 8, 3 },
{ 0, 0, 0, 0, 8, 0, 0 },
{ 6, 8, 0, 5, 3, 0, 0 }
};
struct edge ugraph2_mst[ugraph2_size*ugraph2_size];
for ( int i = 0; i < ( ugraph2_size * ugraph2_size ); i++ )
{
ugraph2_mst[i].u = -1;
ugraph2_mst[i].v = -1;
}
/* arbitrary starting vertex */
int r = 0;
/* get MST for unigraph1, and print */
printf ("Matrix:\n");
ugraph_print ( ugraph1_size, ugraph1 );
printf ("Minimum spanning tree for matrix, from Prim's algorithm:\n");
ugraph_mst_print ( ugraph1_size, mst_prim ( ugraph1_size, ugraph1, r, \
ugraph1_mst ) );
printf ("~~~~~~~~~~~~~~~~~~~~~~~~\n");
/* get MST for unigraph2, and print */
printf ("Matrix:\n");
ugraph_print ( ugraph2_size, ugraph2 );
printf ("Minimum spanning tree for matrix, from Prim's algorithm:\n");
ugraph_mst_print ( ugraph2_size, mst_prim ( ugraph2_size, ugraph2, r, \
ugraph2_mst ) );
printf ("~~~~~~~~~~~~~~~~~~~~~~~~\n");
/* normal return value */
return 0;
}
