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| 1 | +/* Prim's Algorithm |
| 2 | + This is Prim's algorithm to find the lowest cost to traverse it all vertex of a graph or MST(Minimal Spanning Tree) */ |
| 3 | +#include<stdio.h> |
| 4 | +#include<stdlib.h> |
| 5 | +// node type structure |
| 6 | +struct node |
| 7 | +{ |
| 8 | + int no; |
| 9 | + int PI; |
| 10 | + int Key; |
| 11 | + struct node *next; |
| 12 | +}; |
| 13 | +// graph type structure |
| 14 | +struct Graph |
| 15 | +{ |
| 16 | + int V; |
| 17 | + int E; |
| 18 | +}; |
| 19 | +// this function initialize value in Graph structure members |
| 20 | +struct Graph* create_graph(int V,int E) |
| 21 | +{ |
| 22 | + // Graph structure pointer to store value |
| 23 | + struct Graph *G; |
| 24 | + G=(struct Graph*)malloc(sizeof(struct Graph)); |
| 25 | + G->E=E; |
| 26 | + G->V=V; |
| 27 | + return(G); |
| 28 | +} |
| 29 | +// function to check the least priority Key or distance node no. and return it |
| 30 | +int less_priority(struct node *Start) |
| 31 | +{ |
| 32 | + int less=0,no=0; |
| 33 | + less=Start->Key; |
| 34 | + no=Start->no; |
| 35 | + Start=Start->next; |
| 36 | + // until start is not equal to 0 the loop will run |
| 37 | + while(Start!=NULL) |
| 38 | + { |
| 39 | + if(Start->Key<less) |
| 40 | + { |
| 41 | + less=Start->Key; |
| 42 | + no=Start->no; |
| 43 | + } |
| 44 | + Start=Start->next; |
| 45 | + } |
| 46 | + return(no); |
| 47 | +} |
| 48 | +// this function insert the a node |
| 49 | +void insertion(struct node **Start,int no) |
| 50 | +{ |
| 51 | + struct node *temp,*t; |
| 52 | + temp=(struct node*)malloc(sizeof(struct node)); |
| 53 | + temp->no=no; |
| 54 | + temp->PI=0; |
| 55 | + temp->Key=1000; |
| 56 | + temp->next=NULL; |
| 57 | + if(*Start==NULL) |
| 58 | + { temp->Key=0; |
| 59 | + *Start=temp; |
| 60 | + } |
| 61 | + else |
| 62 | + { |
| 63 | + t=*Start; |
| 64 | + while(t->next!=NULL) |
| 65 | + t=t->next; |
| 66 | + t->next=temp; |
| 67 | + } |
| 68 | +} |
| 69 | +// remove node from the queue |
| 70 | +struct node* deletion(struct node **Start) |
| 71 | +{ int check=0; |
| 72 | +struct node *temp,*t,*pre; |
| 73 | + temp=(struct node*)malloc(sizeof(struct node)); |
| 74 | + // finding the less priority node no. |
| 75 | + check=less_priority(*Start); |
| 76 | + // is start no. member is equal to check variable |
| 77 | + if((*Start)->no==check) |
| 78 | + { |
| 79 | + temp->no=(*Start)->no; |
| 80 | + temp->Key=(*Start)->Key; |
| 81 | + temp->PI=(*Start)->PI; |
| 82 | + temp->next=NULL; |
| 83 | + t=*Start; |
| 84 | + *Start=(*Start)->next; |
| 85 | + free(t); |
| 86 | + } |
| 87 | + else |
| 88 | + { t=*Start; |
| 89 | + while(t->next!=NULL) |
| 90 | + { |
| 91 | + pre=t; |
| 92 | + t=t->next; |
| 93 | + if(check==t->no) |
| 94 | + break; |
| 95 | + } |
| 96 | + temp->no=t->no; |
| 97 | + temp->PI=t->PI; |
| 98 | + temp->Key=t->Key; |
| 99 | + temp->next=NULL; |
| 100 | + pre->next=t->next; |
| 101 | + free(t); |
| 102 | + } |
| 103 | + return(temp); |
| 104 | +} |
| 105 | +// this function check whether the particular node is in the queue or not |
| 106 | +int check_in_q(int n,struct node *Start) |
| 107 | +{ |
| 108 | + while(Start!=NULL) |
| 109 | + { |
| 110 | + if(Start->no==n) |
| 111 | + return(1); |
| 112 | + Start=Start->next; |
| 113 | + } |
| 114 | + return(0); |
| 115 | +} |
| 116 | +int get_n(int n,struct node *Start) |
| 117 | +{ int Key; |
| 118 | +// finding the distance of the node no. match with the n(node number) and returning the distance |
| 119 | + while(Start!=NULL) |
| 120 | + { |
| 121 | + if(n==Start->no) |
| 122 | + { |
| 123 | + Key=Start->Key; |
| 124 | + break; |
| 125 | + } |
| 126 | + Start=Start->next; |
| 127 | + } |
| 128 | + return(Key); |
| 129 | +} |
| 130 | +//this function change the Key(distance) and PI(Parent node) value with the given arguments |
| 131 | +void change(int n,struct node **Start,int Key,int PI) |
| 132 | +{ |
| 133 | + struct node *temp; |
| 134 | + if((*Start)->no==n) |
| 135 | + { |
| 136 | + (*Start)->Key=Key; |
| 137 | + (*Start)->PI=PI; |
| 138 | + } |
| 139 | + else |
| 140 | + { |
| 141 | + temp=*Start; |
| 142 | + while(temp!=NULL) |
| 143 | + { |
| 144 | + if(temp->no==n) |
| 145 | + { |
| 146 | + temp->Key=Key; |
| 147 | + temp->PI=PI; |
| 148 | + break; |
| 149 | + } |
| 150 | + temp=temp->next; |
| 151 | + } |
| 152 | + } |
| 153 | +} |
| 154 | +//the main logic to find lowest cost traverse |
| 155 | +void Prims(struct node **Start,struct Graph *G) |
| 156 | +{ int e1,i,n,w; |
| 157 | + struct node *temp; |
| 158 | + int t,cost=0; |
| 159 | + // run until start is not equal to 0 |
| 160 | + while(*Start!=NULL) |
| 161 | + { |
| 162 | + //take the less priority node |
| 163 | + temp=deletion(Start); |
| 164 | + // increase cost |
| 165 | + cost+=temp->Key; |
| 166 | + printf("Enter the no. of nodes connected to V%d",temp->no); |
| 167 | + scanf("%d",&e1); |
| 168 | + for(i=0;i<e1;i++) |
| 169 | + { |
| 170 | + printf("enter the node no of v%d and weight",temp->no); |
| 171 | + scanf("%d%d",&n,&w); |
| 172 | + if(1==check_in_q(n,*Start)) |
| 173 | + { |
| 174 | + t=get_n(n,*Start); |
| 175 | + if(w<t) |
| 176 | + change(n,Start,w,e1); |
| 177 | + } |
| 178 | + } |
| 179 | + } |
| 180 | + printf("\n Cost=%d",cost); |
| 181 | +} |
| 182 | +// driver code |
| 183 | +int main() |
| 184 | +{ |
| 185 | + struct node *Start=NULL; |
| 186 | + struct Graph *G; |
| 187 | + int i,v,e; |
| 188 | + printf("enter vertex and edges"); |
| 189 | + scanf("%d %d",&v,&e); |
| 190 | + G=create_graph(v,e); |
| 191 | + for(i=0;i<G->V;i++) |
| 192 | + insertion(&Start,i+1); |
| 193 | + Prims(&Start,G); |
| 194 | + return 0; |
| 195 | +} |
| 196 | +/* |
| 197 | +Input: |
| 198 | +enter vertex and edges 3 |
| 199 | +6 |
| 200 | +Enter the no. of nodes connected to V1 2 |
| 201 | +enter the node no of v1 and weight2 |
| 202 | +5 |
| 203 | +enter the node no of v1 and weight 3 |
| 204 | +2 |
| 205 | +Enter the no. of nodes connected to V3 2 |
| 206 | +enter the node no of v3 and weight1 |
| 207 | +2 |
| 208 | +enter the node no of v3 and weight 2 |
| 209 | +2 |
| 210 | +Enter the no. of nodes connected to V2 2 |
| 211 | +enter the node no of v2 and weight 1 |
| 212 | +5 |
| 213 | +enter the node no of v2 and weight 3 |
| 214 | +2 |
| 215 | +Output: Cost: 4 |
| 216 | +
|
| 217 | +Time Complexity:O(V^2) |
| 218 | +
|
| 219 | +*/ |
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