Prim's Algorithm in c (#140)
* Added Prim's Algorithm * added Time complexity and algorithm name * added prim's algorithm link * recorrected the link address * added .c exitension * Update time complexity of prim's algo Co-authored-by: Arsenic <54987647+Arsenic-ATG@users.noreply.github.com>pull/146/head
Ujjwal
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@ -4,3 +4,4 @@
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1. [Kruskal Algorithm](c-or-cpp/kruskal-algorithm.cpp)
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2. [Bellman Ford Algorithm](c-or-cpp/bellman-ford.cpp)
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3. [Prim's Algorithm](c-or-cpp/Prim's-algorithm.c)
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/* Prim's Algorithm
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This is Prim's algorithm to find the lowest cost to traverse it all vertex of a graph or MST(Minimal Spanning Tree) */
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#include<stdio.h>
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#include<stdlib.h>
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// node type structure
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struct node
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{
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int no;
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int PI;
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int Key;
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struct node *next;
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};
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// graph type structure
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struct Graph
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{
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int V;
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int E;
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};
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// this function initialize value in Graph structure members
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struct Graph* create_graph(int V,int E)
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{
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// Graph structure pointer to store value
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struct Graph *G;
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G=(struct Graph*)malloc(sizeof(struct Graph));
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G->E=E;
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G->V=V;
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return(G);
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}
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// function to check the least priority Key or distance node no. and return it
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int less_priority(struct node *Start)
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{
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int less=0,no=0;
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less=Start->Key;
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no=Start->no;
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Start=Start->next;
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// until start is not equal to 0 the loop will run
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while(Start!=NULL)
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{
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if(Start->Key<less)
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{
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less=Start->Key;
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no=Start->no;
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}
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Start=Start->next;
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}
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return(no);
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}
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// this function insert the a node
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void insertion(struct node **Start,int no)
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{
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struct node *temp,*t;
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temp=(struct node*)malloc(sizeof(struct node));
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temp->no=no;
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temp->PI=0;
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temp->Key=1000;
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temp->next=NULL;
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if(*Start==NULL)
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{ temp->Key=0;
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*Start=temp;
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}
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else
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{
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t=*Start;
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while(t->next!=NULL)
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t=t->next;
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t->next=temp;
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}
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}
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// remove node from the queue
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struct node* deletion(struct node **Start)
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{ int check=0;
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struct node *temp,*t,*pre;
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temp=(struct node*)malloc(sizeof(struct node));
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// finding the less priority node no.
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check=less_priority(*Start);
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// is start no. member is equal to check variable
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if((*Start)->no==check)
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{
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temp->no=(*Start)->no;
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temp->Key=(*Start)->Key;
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temp->PI=(*Start)->PI;
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temp->next=NULL;
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t=*Start;
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*Start=(*Start)->next;
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free(t);
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}
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else
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{ t=*Start;
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while(t->next!=NULL)
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{
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pre=t;
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t=t->next;
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if(check==t->no)
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break;
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}
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temp->no=t->no;
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temp->PI=t->PI;
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temp->Key=t->Key;
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temp->next=NULL;
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pre->next=t->next;
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free(t);
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}
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return(temp);
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}
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// this function check whether the particular node is in the queue or not
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int check_in_q(int n,struct node *Start)
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{
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while(Start!=NULL)
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{
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if(Start->no==n)
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return(1);
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Start=Start->next;
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}
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return(0);
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}
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int get_n(int n,struct node *Start)
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{ int Key;
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// finding the distance of the node no. match with the n(node number) and returning the distance
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while(Start!=NULL)
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{
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if(n==Start->no)
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{
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Key=Start->Key;
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break;
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}
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Start=Start->next;
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}
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return(Key);
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}
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//this function change the Key(distance) and PI(Parent node) value with the given arguments
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void change(int n,struct node **Start,int Key,int PI)
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{
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struct node *temp;
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if((*Start)->no==n)
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{
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(*Start)->Key=Key;
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(*Start)->PI=PI;
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}
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else
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{
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temp=*Start;
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while(temp!=NULL)
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{
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if(temp->no==n)
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{
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temp->Key=Key;
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temp->PI=PI;
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break;
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}
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temp=temp->next;
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}
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}
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}
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//the main logic to find lowest cost traverse
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void Prims(struct node **Start,struct Graph *G)
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{ int e1,i,n,w;
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struct node *temp;
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int t,cost=0;
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// run until start is not equal to 0
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while(*Start!=NULL)
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{
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//take the less priority node
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temp=deletion(Start);
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// increase cost
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cost+=temp->Key;
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printf("Enter the no. of nodes connected to V%d",temp->no);
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scanf("%d",&e1);
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for(i=0;i<e1;i++)
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{
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printf("enter the node no of v%d and weight",temp->no);
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scanf("%d%d",&n,&w);
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if(1==check_in_q(n,*Start))
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{
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t=get_n(n,*Start);
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if(w<t)
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change(n,Start,w,e1);
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}
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}
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}
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printf("\n Cost=%d",cost);
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}
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// driver code
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int main()
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{
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struct node *Start=NULL;
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struct Graph *G;
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int i,v,e;
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printf("enter vertex and edges");
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scanf("%d %d",&v,&e);
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G=create_graph(v,e);
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for(i=0;i<G->V;i++)
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insertion(&Start,i+1);
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Prims(&Start,G);
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return 0;
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}
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/*
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Input:
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enter vertex and edges 3
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6
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Enter the no. of nodes connected to V1 2
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enter the node no of v1 and weight2
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5
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enter the node no of v1 and weight 3
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2
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Enter the no. of nodes connected to V3 2
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enter the node no of v3 and weight1
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2
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enter the node no of v3 and weight 2
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2
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Enter the no. of nodes connected to V2 2
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enter the node no of v2 and weight 1
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5
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enter the node no of v2 and weight 3
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2
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Output: Cost: 4
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Time Complexity:O(V^2)
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*/
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