Dijkstra's Algorithm Single Source Shortest Path in DSA C

Mental Model

Find the shortest path from one starting point to all other points by always choosing the closest unvisited point next.

Analogy: Imagine you are in a city and want to find the shortest way to every other city by always traveling to the nearest city you haven't visited yet.

Start: S
Graph:
S --2--> A
S --5--> B
A --1--> B
A --3--> C
B --2--> C

Distances:
S: 0
A: ∞
B: ∞
C: ∞

Visited: none

Dry Run Walkthrough

Input: Graph with nodes S, A, B, C and edges: S-A(2), S-B(5), A-B(1), A-C(3), B-C(2); start node S

Goal: Find shortest distances from S to all nodes

Step 1: Initialize distances: S=0, others=∞; mark all unvisited

Distances: S[0], A[∞], B[∞], C[∞]; Visited: none

Why: Set starting point distance to zero and others unknown

Step 2: Pick unvisited node with smallest distance: S(0); visit S

Visited: S; Distances: S[0], A[∞], B[∞], C[∞]

Why: Start from source node

Step 3: Update neighbors of S: A=2, B=5

Distances: S[0], A[2], B[5], C[∞]; Visited: S

Why: Check direct paths from S

Step 4: Pick unvisited node with smallest distance: A(2); visit A

Visited: S, A; Distances: S[0], A[2], B[5], C[∞]

Why: Next closest node to explore

Step 5: Update neighbors of A: B=min(5, 2+1=3)=3, C=2+3=5

Distances: S[0], A[2], B[3], C[5]; Visited: S, A

Why: Found shorter path to B and new path to C

Step 6: Pick unvisited node with smallest distance: B(3); visit B

Visited: S, A, B; Distances: S[0], A[2], B[3], C[5]

Why: Next closest node to explore

Step 7: Update neighbors of B: C=min(5, 3+2=5)=5 (no change)

Distances: S[0], A[2], B[3], C[5]; Visited: S, A, B

Why: No shorter path to C found

Step 8: Pick unvisited node with smallest distance: C(5); visit C

Visited: S, A, B, C; Distances: S[0], A[2], B[3], C[5]

Why: All nodes visited, algorithm ends

Result:

Final distances: S[0] -> A[2] -> B[3] -> C[5]

Annotated Code

DSA C

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

#define V 4

int minDistance(int dist[], bool sptSet[]) {
    int min = INT_MAX, min_index = -1;
    for (int v = 0; v < V; v++) {
        if (!sptSet[v] && dist[v] < min) {
            min = dist[v];
            min_index = v;
        }
    }
    return min_index;
}

void dijkstra(int graph[V][V], int src) {
    int dist[V];
    bool sptSet[V];

    for (int i = 0; i < V; i++) {
        dist[i] = INT_MAX;
        sptSet[i] = false;
    }

    dist[src] = 0;

    for (int count = 0; count < V - 1; count++) {
        int u = minDistance(dist, sptSet); // pick closest unvisited
        sptSet[u] = true; // mark visited

        for (int v = 0; v < V; v++) {
            if (!sptSet[v] && graph[u][v] && dist[u] != INT_MAX && dist[u] + graph[u][v] < dist[v]) {
                dist[v] = dist[u] + graph[u][v]; // update distance
            }
        }
    }

    printf("Vertex Distance from Source\n");
    for (int i = 0; i < V; i++) {
        printf("%c -> %d\n", 'S' + i, dist[i]);
    }
}

int main() {
    int graph[V][V] = {
        {0, 2, 5, 0},
        {0, 0, 1, 3},
        {0, 0, 0, 2},
        {0, 0, 0, 0}
    };

    dijkstra(graph, 0); // source is S (index 0)
    return 0;
}

int u = minDistance(dist, sptSet); // pick closest unvisited

select the unvisited node with the smallest current distance

sptSet[u] = true; // mark visited

mark this node as visited so we don't revisit

if (!sptSet[v] && graph[u][v] && dist[u] != INT_MAX && dist[u] + graph[u][v] < dist[v]) {

check if neighbor v can be reached with a shorter path through u

dist[v] = dist[u] + graph[u][v]; // update distance

update distance to neighbor v with shorter path found

Output

Vertex Distance from Source

S -> 0

A -> 2

B -> 3

C -> 5

Complexity Analysis

Time: O(V^2) because we check all vertices to find the minimum distance and update neighbors in nested loops

Space: O(V) for distance and visited arrays to store shortest path info

vs Alternative: Compared to using a priority queue (min-heap), this simple approach is slower (O(V^2) vs O(E log V)) but easier to understand and implement for small graphs

Edge Cases

Graph with no edges

Distances remain infinite except source which is zero

DSA C

dist[src] = 0;

Graph with single node

Distance to itself is zero, algorithm ends immediately

DSA C

for (int count = 0; count < V - 1; count++) {

Disconnected nodes

Distances to unreachable nodes remain infinite (INT_MAX)

DSA C

if (!sptSet[v] && graph[u][v] && dist[u] != INT_MAX && dist[u] + graph[u][v] < dist[v]) {

When to Use This Pattern

When you need shortest paths from one point to all others in a weighted graph without negative edges, reach for Dijkstra's algorithm because it efficiently finds minimum distances step-by-step.

Common Mistakes

Mistake: Not marking nodes as visited after selecting them
Fix: Set sptSet[u] = true immediately after picking the node with minimum distance

Mistake: Updating distances without checking if current distance is INT_MAX
Fix: Add condition dist[u] != INT_MAX before updating neighbors to avoid overflow

Summary

Finds shortest paths from a single source to all nodes in a weighted graph with non-negative edges.

Use when you want minimum travel cost or distance from one point to many points.

Always pick the closest unvisited node next and update neighbors to find shortest paths stepwise.