Add comments to the code and display outputInt main(void)//Input array containing edges of the graph//(X,Y,W)tuple represents an edge from 'x' to y' having weight w'struct Edge edges I={0,1,4.).(1,2,4).(2,0,4),(2,1,4)int n=size of (edges)/size of (edges|0]);struct Graph *graph=createGraph(edges,n);return 0;static int minsum=100000;static voidgetMinPathSum(Dictionary<int,ArrayList>graph,bool[visited,ArrayLinecessary,int source,int dest,int currsum)if(src==dest)bool flag=true;for each(int i in necessary)if(!visitedi])flag=false;break:if (flag )minsum=Math. Min(minsum,currsum);return;elselvisited[src]=true;for(auto node:graph[src])if (Ivisited[node.first])visited [node.first])=true;getMinPathSum(graph, visited,necessary,node.first,dest,currsum+node.second);visited [node.first]=false; First we have to create a graph with all nodes and all edges adweights are same using the following code:Int main(void){//Input array containing edges of the graph//(X,Y,W)tuple represents an edge from 'x' to 'y' having weight 'w'struct Edge edges ]={{0,1,4,),(1,2,4},(2,0,4),(2,1,4)};int n=size of (edges)/size of (edges[0]);struct Graph *graph3createGraph(edges,n);return 0;static int minsum3D100000;static voidgetMinPathSum(Dictionarygraph,bool]visited,ArrayLinecessary,int source,int dest,int currsum){if(src==dest){bool flag=true%;for each(int i in necessary){if(!visited[i]){flag=false;break;}if(flag)minsum=Math.Min(minsum,currsum);return;else{visited[src)=true;for(auto node:graph[srcl){if(!visited[node.first){visited [node.first])=true;getMinPathSum(graph, visited,necessary,node.first,dest,currsum+node.second);visited [node.first]=false;}

#include <stdio.h>#include <stdlib.h> // Define the maximum number of vertices in the…

Answered: Add comments to the code and display…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

See similar textbooks

Related questions

Question

Add comments to the code and display output
Int main(void)
//Input array containing edges of the graph
//(X,Y,W)tuple represents an edge from 'x' to y' having weight w'
struct Edge edges I=

{0,1,4.).(1,2,4).(2,0,4),(2,1,4)
int n=size of (edges)/size of (edges|0]);
struct Graph *graph=createGraph(edges,n);
return 0;

static int minsum=100000;
static void
getMinPathSum(Dictionary<int,ArrayList>graph,bool[visited,ArrayLi
necessary,int source,int dest,int currsum)
if(src==dest)

bool flag=true;
for each(int i in necessary)
if(!visitedi])
flag=false;
break:

if (flag )
minsum=Math. Min(minsum,currsum);
return;
elsel
visited[src]=true;
for(auto node:graph[src])

if (Ivisited[node.first])
visited [node.first])=true;
getMinPathSum(graph, visited,
necessary,node.first,dest,currsum+node.second);
visited [node.first]=false;

First we have to create a graph with all nodes and all edges ad
weights are same using the following code:
Int main(void)
{
//Input array containing edges of the graph
//(X,Y,W)tuple represents an edge from 'x' to 'y' having weight 'w'
struct Edge edges ]=
{
{0,1,4,),(1,2,4},(2,0,4),(2,1,4)
};
int n=size of (edges)/size of (edges[0]);
struct Graph *graph3createGraph(edges,n);
return 0;
static int minsum3D100000;
static void
getMinPathSum(Dictionary<int,ArrayList>graph,bool]visited,ArrayLi
necessary,int source,int dest,int currsum)
{
if(src==dest)
{
bool flag=true%;
for each(int i in necessary)
{
if(!visited[i])
{
flag=false;
break;
}
if(flag)
minsum=Math.Min(minsum,currsum);
return;
else{
visited[src)=true;
for(auto node:graph[srcl)
{
if(!visited[node.first)
{
visited [node.first])=true;
getMinPathSum(graph, visited,
necessary,node.first,dest,currsum+node.second);
visited [node.first]=false;
}

Expert Solution

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, computer-science and related others by exploring similar questions and additional content below.

Similar questions