HUFFMAN CODING (Mini Project) - Data Structures Source Code in C

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Home » C Home » Data Structures Home » HUFFMAN CODING (Mini Project)

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Title	HUFFMAN CODING (Mini Project)
Author	G.KALAIARASU
Author Email	kalaiarasutech [at] gmail.com
Description	This program is very useful for engineers those who have been taking Information coding techniques as major.
Category	C » Data Structures
Hits	417569
Code	Select and Copy the Code
Code : #include<iostream.h> #include<string.h> #include<math.h> #include<stdlib.h> #include<conio.h> struct tree { char a[20]; int s; struct tree left,right; }root=NULL,tt[20]={NULL},temp,temp2,t2,ri,le; struct pqu { int info; char a[20]; struct pqu ptr; }front=NULL,t,par,t1,p1,p2; struct pqu* fp(int info) { struct pqu p=NULL; for(t1=front;t1->info<info&&t1!=NULL;t1=t1->ptr) { p=t1; } return (p); } void enqu(char a[20],int p) { t=(struct pqu)malloc(sizeof(struct pqu)); strcpy(t->a,a); t->info=p; t->ptr=NULL; if(front==NULL) { front=t; } else { par=fp(p); if(par==NULL) { t->ptr=front; front=t; } else { t->ptr=par->ptr; par->ptr=t; } } } struct pqu* dequ() { t1=front; front=front->ptr; return t1; } void info(char c[2]) { int m=0,i; temp2=root; while(strcmp(c,temp2->a)!=0) { t2=temp2->left; for(i=0;i<strlen(t2->a);i++) { if(t2->a[i]==c[0]) { temp2=temp2->left; m=1; cout<<"0"; break; } } if(m!=1) { temp2=temp2->right; cout<<1; } m=0; } } void insert() { char a1[20],b1[20],v1[20]; int i,j,z=0,l; while(front!=NULL) { p1=dequ(); strcpy(a1,p1->a); l=p1->info; p2=dequ(); if(p2==NULL) break; strcpy(b1,p2->a); strcpy(v1,a1); temp=(struct tree)malloc(sizeof(struct tree)); strcpy(temp->a,strcat(v1,b1)); temp->s=l+p2->info; temp->left=NULL; temp->right=NULL; temp2=temp; root=temp; for(i=0;i<z;) { if(strcmp(tt[i]->a,a1)==0) { temp->left=tt[i]; for(l=i;l<z;l++) { tt[l]=tt[l+1]; } i=0; continue; } else if(strcmp(tt[i]->a,b1)==0) { temp->right=tt[i]; for(l=i;l<z;l++) { tt[l]=tt[l+1]; } i=0; continue; } i++; } if(temp->left==NULL) { le=(struct tree)malloc(sizeof(struct tree)); strcpy(le->a,a1); le->left=NULL; le->right=NULL; temp2->left=le; } if(temp->right==NULL) { ri=(struct tree)malloc(sizeof(struct tree)); strcpy(ri->a,b1); ri->left=NULL; ri->right=NULL; temp2->right=ri; } if(front!=NULL) enqu(temp2->a,temp2->s); tt[z++]=temp2; } } void disp(struct tree rt) { if(rt!=NULL) { disp(rt->left); cout<<" "<<rt->a; disp(rt->right); } } void main() { textmode(MONO); int i=0,g,h,p,y,n; char m[20],b[20][2],re; while(1) { clrscr(); cout<<"=================================================================== ========"; cout<<" HUFFMAN CODING "; cout<<"=================================================================== ======== "; cout<<"Enter the total no of characters : "; cin>>n; for(i=0;i<n;i++) { cout<<" Enter the character : "; cin>>m; strcpy(b[i],m); cout<<" Enter frequency for "<<m<<" : "; cin>>g; enqu(m,g); } insert(); disp(root); clrscr(); cout<<"=================================================================== ============"; cout<<" The corresponding codes are..... "; cout<<"=================================================================== ============"; for(i=0;i<n;i++) { cout<<" "<<b[i]<<" : "; info(b[i]); } cout<<" DO YOU WANT TO CONTINUE Y OR N: "; cin>>re; if(re=='y'\|\|re=='Y') continue; else exit(0); } }

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