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@ajinkyajawale14499
Created August 6, 2019 08:07
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Dijkstra Shortest path algorithm using greedy method
#include<iostream>
#include<bits/stdc++.h>
#define V 10
#define INT_MAX 99999999
using namespace std;
int mindistance(int dist[],bool sptset[])
{
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;
}
void printdijkstra(int dist[V],int n)
{
for(int i=0;i<n;i++)
printf("%d tt %d \n",i,dist[i] );
}
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 be 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 the 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]
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
printdijkstra(dist, V);
}
}
int main()
{
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, 14, 10, 0, 2, 0, 0},
{0, 0, 0, 0, 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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