An implementation of Dijkstra's algorithm is given below.
#include
#include
#include "debug.h"
#define NUM_NODES 8
#define NONE 9999
#define X(l) ((l) - 'A')
#define Y(n) ((n) + 'A')
struct _NODE
{
int iDist;
int iPrev;
};
typedef struct _NODE NODE;
struct _QITEM
{
int iNode;
int iDist;
int iPrev;
struct _QITEM *qNext;
};
typedef struct _QITEM QITEM;
QITEM *qHead = NULL;
/*
H
/ \
1 2
/ \
F---4---G
/ \
4 2
/ \
E---2---A---3---D
\ / \ /
3 1 2 3
\ / \ /
B---2---C
*/
int AdjMatrix[NUM_NODES][NUM_NODES] =
{
/* A B C D E F G H */
/* A */ { NONE, 1, 2, 3, 2, NONE, NONE, NONE },
/* B */ { 1, NONE, 2, NONE, 3, NONE, NONE, NONE },
/* C */ { 2, 2, NONE, 2, NONE, NONE, NONE, NONE },
/* D */ { 3, NONE, 2, NONE, NONE, NONE, 2, NONE },
/* E */ { 2, 3, NONE, NONE, NONE, 4, NONE, NONE },
/* F */ { NONE, NONE, NONE, NONE, 4, NONE, 4, 1 },
/* G */ { NONE, NONE, NONE, 2, NONE, 4, NONE, 2 },
/* H */ { NONE, NONE, NONE, NONE, NONE, 1, 2, NONE }
};
int g_qCount = 0;
void print_path (NODE *rgnNodes, int chNode)
{
if (rgnNodes[chNode].iPrev != NONE)
{
print_path(rgnNodes, rgnNodes[chNode].iPrev);
}
printf (" %c", Y(chNode));
fflush(stdout);
}
void enqueue (int iNode, int iDist, int iPrev)
{
QITEM *qNew = (QITEM *) malloc(sizeof(QITEM));
QITEM *qLast = qHead;
if (!qNew)
{
fprintf(stderr, "Out of memory.\n");
exit(1);
}
qNew->iNode = iNode;
qNew->iDist = iDist;
qNew->iPrev = iPrev;
qNew->qNext = NULL;
if (!qLast)
{
qHead = qNew;
}
else
{
while (qLast->qNext) qLast = qLast->qNext;
qLast->qNext = qNew;
}
g_qCount++;
ASSERT(g_qCount);
}
void dequeue (int *piNode, int *piDist, int *piPrev)
{
QITEM *qKill = qHead;
if (qHead)
{
ASSERT(g_qCount);
*piNode = qHead->iNode;
*piDist = qHead->iDist;
*piPrev = qHead->iPrev;
qHead = qHead->qNext;
free(qKill);
g_qCount--;
}
}
int qcount (void)
{
return(g_qCount);
}
int main(void)
{
NODE rgnNodes[NUM_NODES];
char rgchLine[255];
char chStart, chEnd, ch;
int iPrev, iNode;
int i, iCost, iDist;
for (ch = 'A'; ch <= 'A' + NUM_NODES; ch++)
{
rgnNodes[X(ch)].iDist = NONE;
rgnNodes[X(ch)].iPrev = NONE;
}
printf("What is the starting node? ");
gets(rgchLine);
chStart = toupper(rgchLine[0]);
printf("What is the ending node? ");
gets(rgchLine);
chEnd = toupper(rgchLine[0]);
if (chStart == chEnd)
{
printf("Shortest path is 0 in cost.\nJust stay where you are.\n");
exit(0);
}
else
{
chStart = X(chStart);
chEnd = X(chEnd);
rgnNodes[chStart].iDist = 0;
rgnNodes[chStart].iPrev = NONE;
enqueue (chStart, 0, NONE);
while (qcount() > 0)
{
dequeue (&iNode, &iDist, &iPrev);
for (i = 0; i < NUM_NODES; i++)
{
if ((iCost = AdjMatrix[iNode][i]) != NONE)
{
if ((NONE == rgnNodes[i].iDist) ||
(rgnNodes[i].iDist > (iCost + iDist)))
{
rgnNodes[i].iDist = iDist + iCost;
rgnNodes[i].iPrev = iNode;
enqueue (i, iDist + iCost, iNode);
}
}
}
}
printf("Shortest path is %d in cost.\n", rgnNodes[chEnd].iDist);
printf("Path is: ");
print_path(rgnNodes, chEnd);
}
exit(0);
}
|