Hel - 1 year ago 89

C Question

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`

`#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;

}

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Answer Source

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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