We recommend to read following two posts as a prerequisite of this post.
We have discussed Dijkstra’s algorithm and its implementation for adjacency matrix representation of graphs. The time complexity for the matrix representation is O(V^2). In this post, O(ELogV) algorithm for adjacency list representation is discussed.
As discussed in the previous post, in Dijkstra’s algorithm, two sets are maintained, one set contains list of vertices already included in SPT (Shortest Path Tree), other set contains vertices not yet included. With adjacency list representation, all vertices of a graph can be traversed in O(V+E) time using BFS. The idea is to traverse all vertices of graph using BFS and use a Min Heap to store the vertices not yet included in SPT (or the vertices for which shortest distance is not finalized yet). Min Heap is used as a priority queue to get the minimum distance vertex from set of not yet included vertices. Time complexity of operations like extract-min and decrease-key value is O(LogV) for Min Heap.
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Following are the detailed steps.
1) Create a Min Heap of size V where V is the number of vertices in the given graph. Every node of min heap contains vertex number and distance value of the vertex.
2) Initialize Min Heap with source vertex as root (the distance value assigned to source vertex is 0). The distance value assigned to all other vertices is INF (infinite).
3) While Min Heap is not empty, do following
…..a) Extract the vertex with minimum distance value node from Min Heap. Let the extracted vertex be u.
…..b) For every adjacent vertex v of u, check if v is in Min Heap. If v is in Min Heap and distance value is more than weight of u-v plus distance value of u, then update the distance value of v.
Let us understand with the following example. Let the given source vertex be 0

Initially, distance value of source vertex is 0 and INF (infinite) for all other vertices. So source vertex is extracted from Min Heap and distance values of vertices adjacent to 0 (1 and 7) are updated. Min Heap contains all vertices except vertex 0.
The vertices in green color are the vertices for which minimum distances are finalized and are not in Min Heap
Since distance value of vertex 1 is minimum among all nodes in Min Heap, it is extracted from Min Heap and distance values of vertices adjacent to 1 are updated (distance is updated if the a vertex is not in Min Heap and distance through 1 is shorter than the previous distance). Min Heap contains all vertices except vertex 0 and 1.

Pick the vertex with minimum distance value from min heap. Vertex 7 is picked. So min heap now contains all vertices except 0, 1 and 7. Update the distance values of adjacent vertices of 7. The distance value of vertex 6 and 8 becomes finite (15 and 9 respectively).

Pick the vertex with minimum distance from min heap. Vertex 6 is picked. So min heap now contains all vertices except 0, 1, 7 and 6. Update the distance values of adjacent vertices of 6. The distance value of vertex 5 and 8 are updated.
Above steps are repeated till min heap doesn’t become empty. Finally, we get the following shortest path tree.

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
struct AdjListNode
{
int dest;
int weight;
struct AdjListNode* next;
};
struct AdjList
{
struct AdjListNode *head;
};
struct Graph
{
int V;
struct AdjList* array;
};
struct AdjListNode* newAdjListNode(int dest, int weight)
{
struct AdjListNode* newNode =
(struct AdjListNode*) malloc(sizeof(struct AdjListNode));
newNode->dest = dest;
newNode->weight = weight;
newNode->next = NULL;
return newNode;
}
struct Graph* createGraph(int V)
{
struct Graph* graph = (struct Graph*) malloc(sizeof(struct Graph));
graph->V = V;
graph->array = (struct AdjList*) malloc(V * sizeof(struct AdjList));
for (int i = 0; i < V; ++i)
graph->array[i].head = NULL;
return graph;
}
void addEdge(struct Graph* graph, int src, int dest, int weight)
{
struct AdjListNode* newNode = newAdjListNode(dest, weight);
newNode->next = graph->array[src].head;
graph->array[src].head = newNode;
newNode = newAdjListNode(src, weight);
newNode->next = graph->array[dest].head;
graph->array[dest].head = newNode;
}
struct MinHeapNode
{
int v;
int dist;
};
struct MinHeap
{
int size;
int capacity;
int *pos;
struct MinHeapNode **array;
};
struct MinHeapNode* newMinHeapNode(int v, int dist)
{
struct MinHeapNode* minHeapNode =
(struct MinHeapNode*) malloc(sizeof(struct MinHeapNode));
minHeapNode->v = v;
minHeapNode->dist = dist;
return minHeapNode;
}
struct MinHeap* createMinHeap(int capacity)
{
struct MinHeap* minHeap =
(struct MinHeap*) malloc(sizeof(struct MinHeap));
minHeap->pos = (int *)malloc(capacity * sizeof(int));
minHeap->size = 0;
minHeap->capacity = capacity;
minHeap->array =
(struct MinHeapNode**) malloc(capacity * sizeof(struct MinHeapNode*));
return minHeap;
}
void swapMinHeapNode(struct MinHeapNode** a, struct MinHeapNode** b)
{
struct MinHeapNode* t = *a;
*a = *b;
*b = t;
}
void minHeapify(struct MinHeap* minHeap, int idx)
{
int smallest, left, right;
smallest = idx;
left = 2 * idx + 1;
right = 2 * idx + 2;
if (left < minHeap->size &&
minHeap->array[left]->dist < minHeap->array[smallest]->dist )
smallest = left;
if (right < minHeap->size &&
minHeap->array[right]->dist < minHeap->array[smallest]->dist )
smallest = right;
if (smallest != idx)
{
MinHeapNode *smallestNode = minHeap->array[smallest];
MinHeapNode *idxNode = minHeap->array[idx];
minHeap->pos[smallestNode->v] = idx;
minHeap->pos[idxNode->v] = smallest;
swapMinHeapNode(&minHeap->array[smallest], &minHeap->array[idx]);
minHeapify(minHeap, smallest);
}
}
int isEmpty(struct MinHeap* minHeap)
{
return minHeap->size == 0;
}
struct MinHeapNode* extractMin(struct MinHeap* minHeap)
{
if (isEmpty(minHeap))
return NULL;
struct MinHeapNode* root = minHeap->array[0];
struct MinHeapNode* lastNode = minHeap->array[minHeap->size - 1];
minHeap->array[0] = lastNode;
minHeap->pos[root->v] = minHeap->size-1;
minHeap->pos[lastNode->v] = 0;
--minHeap->size;
minHeapify(minHeap, 0);
return root;
}
void decreaseKey(struct MinHeap* minHeap, int v, int dist)
{
int i = minHeap->pos[v];
minHeap->array[i]->dist = dist;
while (i && minHeap->array[i]->dist < minHeap->array[(i - 1) / 2]->dist)
{
minHeap->pos[minHeap->array[i]->v] = (i-1)/2;
minHeap->pos[minHeap->array[(i-1)/2]->v] = i;
swapMinHeapNode(&minHeap->array[i], &minHeap->array[(i - 1) / 2]);
i = (i - 1) / 2;
}
}
bool isInMinHeap(struct MinHeap *minHeap, int v)
{
if (minHeap->pos[v] < minHeap->size)
return true;
return false;
}
void printArr(int dist[], int n)
{
printf("Vertex Distance from Source\n");
for (int i = 0; i < n; ++i)
printf("%d \t\t %d\n", i, dist[i]);
}
void dijkstra(struct Graph* graph, int src)
{
int V = graph->V;
int dist[V];
struct MinHeap* minHeap = createMinHeap(V);
for (int v = 0; v < V; ++v)
{
dist[v] = INT_MAX;
minHeap->array[v] = newMinHeapNode(v, dist[v]);
minHeap->pos[v] = v;
}
minHeap->array[src] = newMinHeapNode(src, dist[src]);
minHeap->pos[src] = src;
dist[src] = 0;
decreaseKey(minHeap, src, dist[src]);
minHeap->size = V;
while (!isEmpty(minHeap))
{
struct MinHeapNode* minHeapNode = extractMin(minHeap);
int u = minHeapNode->v;
struct AdjListNode* pCrawl = graph->array[u].head;
while (pCrawl != NULL)
{
int v = pCrawl->dest;
if (isInMinHeap(minHeap, v) && dist[u] != INT_MAX &&
pCrawl->weight + dist[u] < dist[v])
{
dist[v] = dist[u] + pCrawl->weight;
decreaseKey(minHeap, v, dist[v]);
}
pCrawl = pCrawl->next;
}
}
printArr(dist, V);
}
int main()
{
int V = 9;
struct Graph* graph = createGraph(V);
addEdge(graph, 0, 1, 4);
addEdge(graph, 0, 7, 8);
addEdge(graph, 1, 2, 8);
addEdge(graph, 1, 7, 11);
addEdge(graph, 2, 3, 7);
addEdge(graph, 2, 8, 2);
addEdge(graph, 2, 5, 4);
addEdge(graph, 3, 4, 9);
addEdge(graph, 3, 5, 14);
addEdge(graph, 4, 5, 10);
addEdge(graph, 5, 6, 2);
addEdge(graph, 6, 7, 1);
addEdge(graph, 6, 8, 6);
addEdge(graph, 7, 8, 7);
dijkstra(graph, 0);
return 0;
}
Output:
Vertex Distance from Source
0 0
1 4
2 12
3 19
4 21
5 11
6 9
7 8
8 14
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Time Complexity: The time complexity of the above code/algorithm looks O(V^2) as there are two nested while loops. If we take a closer look, we can observe that the statements in inner loop are executed O(V+E) times (similar to BFS). The inner loop has decreaseKey() operation which takes O(LogV) time. So overall time complexity is O(E+V)*O(LogV) which is O((E+V)*LogV) = O(ELogV)
Note that the above code uses Binary Heap for Priority Queue implementation. Time complexity can be reduced to O(E + VLogV) using Fibonacci Heap. The reason is, Fibonacci Heap takes O(1) time for decrease-key operation while Binary Heap takes O(Logn) time.
Notes:
1) The code calculates shortest distance, but doesn’t calculate the path information. We can create a parent array, update the parent array when distance is updated (like prim’s implementation) and use it show the shortest path from source to different vertices.
2) The code is for undirected graph, same dijekstra function can be used for directed graphs also.
3) The code finds shortest distances from source to all vertices. If we are interested only in shortest distance from source to a single target, we can break the for loop when the picked minimum distance vertex is equal to target (Step 3.a of algorithm).
4) Dijkstra’s algorithm doesn’t work for graphs with negative weight edges. For graphs with negative weight edges, Bellman–Ford algorithm can be used, we will soon be discussing it as a separate post.
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