Hel Hel - 3 months ago 26
C Question

how to understand details in this program of Dijkstra’s shortest path algorithm

I'm reading Dijkstra’s shortest path algorithm .


Given a graph and a source vertex in graph, find shortest paths value from
source to all vertices in the given graph.


I can't understand why
dist[u] != INT_MAX
is needed.

#include <stdio.h>
#include <limits.h>

// Number of vertices in the graph
#define V 9

// A utility function to find the vertex with minimum distance value, from
// the set of vertices not yet included in shortest path tree
int minDistance(int dist[], bool sptSet[])
{
// Initialize min value
int min = INT_MAX, min_index;

for (int v = 0; v < V; v++)
if (sptSet[v] == false && dist[v] <= min)
min = dist[v], min_index = v;

return min_index;
}

// A utility function to print the constructed distance array
void printSolution(int dist[], int n)
{
printf("Vertex Distance from Source\n");
for (int i = 0; i < V; i++)
printf("%d \t\t %d\n", i, dist[i]);
}

// Funtion that implements Dijkstra's single source shortest path algorithm
// for a graph represented using adjacency matrix representation
void dijkstra(int graph[V][V], int src)
{
int dist[V]; // The output array. dist[i] will hold the shortest
// distance from src to i

bool sptSet[V]; // sptSet[i] will true if vertex i is included in shortest
// path tree or shortest distance from src to i is finalized

// Initialize all distances as INFINITE and stpSet[] as false
for (int i = 0; i < V; i++)
dist[i] = INT_MAX, sptSet[i] = false;

// Distance of source vertex from itself is always 0
dist[src] = 0;

// Find shortest path for all vertices
for (int count = 0; count < V-1; count++)
{
// Pick the minimum distance vertex from the set of vertices not
// yet processed. u is always equal to src in first iteration.
int u = minDistance(dist, sptSet);

// Mark the picked vertex as processed
sptSet[u] = true;

// Update dist value of the adjacent vertices of the picked vertex.
for (int v = 0; v < V; v++)

// Update dist[v] only if is not in sptSet, there is an edge from
// u to v, and total weight of path from src to v through u is
// smaller than current value of dist[v]

/*
*
* Why dist[u] != INT_MAX is needed? Can this condition be deleted?
*
*/
if (!sptSet[v] && graph[u][v] && dist[u] != INT_MAX
&& dist[u]+graph[u][v] < dist[v])
dist[v] = dist[u] + graph[u][v];
}

// print the constructed distance array
printSolution(dist, V);
}

int main(void)
{
/* Let us create the example graph discussed above */
int graph[V][V] = {{0, 4, 0, 0, 0, 0, 0, 8, 0},
{4, 0, 8, 0, 0, 0, 0, 11, 0},
{0, 8, 0, 7, 0, 4, 0, 0, 2},
{0, 0, 7, 0, 9, 14, 0, 0, 0},
{0, 0, 0, 9, 0, 10, 0, 0, 0},
{0, 0, 4, 0, 10, 0, 2, 0, 0},
{0, 0, 0, 14, 0, 2, 0, 1, 6},
{8, 11, 0, 0, 0, 0, 1, 0, 7},
{0, 0, 2, 0, 0, 0, 6, 7, 0}
};

dijkstra(graph, 0);

return 0;
}

Answer

Without the check dist[u] != INT_MAX, dist[u]+graph[u][v] may cause signed integer overflow, which invokes undefined behavior.

With the check, thanks to short-circuit evaluation, dist[u]+graph[u][v] < dist[v] won't be evaluated when the value in dist[u] is INT_MAX, so the undefined behavior can be avoided if the input costs are small enough not to cause overflow.

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