Databases and Data Structures [Individual Assignment]

8/8/2019 Databases and Data Structures [Individual Assignment]

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Asia Pacific Institute of Information Technology

Databases and Data Structures AAPP001-3-2

Cover Page

Group Project

Student declaration:

I declare that:

yWe understand what is meant by plagiarismyThe implication of plagiarism have been explained to us by our lectureryThis project is all our work and I have acknowledged any use of the published or

unpublished works of other people.Group Leaders Signature:

Date:17/09/2009

Total number of pages including this cover

page: Class Code: DF 0931 ICT

Submission

Date:July 29 ,2009 Due Date: September 17, 2009

Students'

Full Names

Leader:

Praveena Sarathchandra (CB003403)

Members:

Pulasthi Perera (CB003472)

Total number of pages including this cover page:


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Table of Contents

Acknowledgement .................................................................................................................... 3

Introduction .............................................................................................................................. 4

Project Description .................................................................................................................... 5

Implementation of Link Lists .................................................................................................. 5

Implementation of Circular Queue......................................................................................... 6

Critical Evaluation......................................................................... Error! Bookmark not defined.

Appendices ............................................................................................................................... 7

Sourse code .............................................................................. Error! Bookmark not defined.

Bibliography ........................................................................................................................ 19


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Acknowledgement

First of all we sincerely thank our lecturer Mr. Udesh Amarasinghe for his continuous

guidance and support in the preparation of this report. With the thorough guidance andexpertise that we gained from him we are able to properly get a clear idea of what we

had to do in this project.

We also would like contribute our thanks to the computer laboratories and the library of

the Asia Pacific Institute of Information Technology, for basically providing us with

necessary Internet facilities and reference facilities for this project.

We sincerely thank our friends who had always been around to support us. And alsowould like to express or sincere gratitude for all people who gave their immense support

to make this project success.

Last but not least we have to thankful for our family members who make us free from allthe duties, which help us to finish this project on time.


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Introduction

This project had been taken out as an objective of completion of the project assigned for

the module Database and Data Structures, According to the requirements highlighted in

the project. The two scenarios covered in the project were covered accordingly using the

knowledge gained throughout the module.

We were implementing and develop the program successfully during the time period.

The first part of the project was implementing a program to handle the emergency room

in a hospital using C programming language. This part check the knowledge of link lists

in the module and the next part was covered using the knowledge of circular queues. In

here we have to implement and develop the program to handle paki9ng of vehicles in a

car park.


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Project Description

Implementation of Link Lists

The project is based on a patient scheduling system of an emergency room of a hospital.I had to use linked lists in C programing language for implementation.

In this program there are six options in the main menu. They are to add a new patient tothe end of the list, add patients to the beginning of the list, move patient to front, printthe entire list, and Remove a node from the list and to exit from the program.

The program has to be user friendly and user can be able to use the functions of thisprogram easily without any complexity.


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Implementation of Circular Queue

In this task we where assign to make a car park system with a pre define size of 10.

The main idea is to keep track of all vehicles which are two types vans and cars so thatnecessary billing can be done

What we have done is a menu driven application which as all the functions such asadding, remove, delete and calculating billing are one of the main functions available.

As this is a queue the counter the increase when a vehicle is removed.

In case user miss enters the license number to re-edit it without removing we haveincluded an addition function to the system where the user enters the miss enteredlicense number and replace it by entering the real license number.


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Appendices

Source Codes

Linked Lists

#include

#include

#include

#include

#include

int id=100;

struct list{

int id;

char name[20];

int age;

int cstate;

char ail[20];

struct list *next;

}*head;

void menu();

void initialize();

void addPatientF();

void addPatientT();


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void preempt();

void display();

void removenode(struct list *head);

main(){

int choice;

struct list *head;

head = NULL;

initialize();

do{

menu();

scanf("%d", &choice);

switch(choice){

case 1:

addPatientT();

break;

case 2:

addPatientF();

break;


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case 3:

preempt();

break;

case 4:

display();

break;

case 5:

removenode(&head);

break;

case 6:

exit(0);

break;

default:

printf("Wrong input");

getch();

}

}while(choice != 5);

if(choice ==5){

exit(1);


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}

}

void initialize(){

head = NULL;

}

void addPatientF(){

struct list *newnode;

char name[20];

int age;

int cstate;

char ail[20];

clrscr();

printf("Enter New Patient details\n\n");

printf("ID: %d", id++);

printf("\n--------Enter new patient details---------\n\n");

printf("\tName: ");

fflush(stdin);


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gets(name);

printf("\n\tAge: ");

fflush(stdin);

scanf("%d", &age);

printf("\n\tCritical State: (Rate 0-3)");

fflush(stdin);

scanf("%d", &cstate);

printf("\n\tAilment: ");

fflush(stdin);

gets(ail);

newnode = (struct list*)malloc(sizeof(struct list));

newnode->id = id;

strcpy(newnode->name, name);

newnode->age = age;

newnode->cstate = cstate;

strcpy(newnode->ail, ail);

if(head == NULL){

head = newnode;

head->next = NULL;


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} else {

newnode->next = head;

head = newnode;

}

}

void addPatientT(){

struct list *newnode;

struct list *tmp = head;

char name[20];

int age;

int cstate;

char ail[20];

clrscr();

printf("Enter New Patient details\n\n");

printf("ID: %d", id++);

printf("\n--------Enter new patient details---------\n\n");

printf("\tName: ");


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fflush(stdin);

gets(name);

printf("\n\tAge: ");

fflush(stdin);

scanf("%d", &age);

printf("\n\tCritical State: (Rate 0-3)");

fflush(stdin);

scanf("%d", &cstate);

printf("\n\tAilment: ");

fflush(stdin);

gets(ail);

newnode = (struct list*)malloc(sizeof(struct list));

newnode->id = id;

strcpy(newnode->name, name);

newnode->age = age;

newnode->cstate = cstate;

strcpy(newnode->ail, ail);

newnode->next=NULL;

if(head == NULL){


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head = newnode;

head->next = NULL;

} else {

newnode->next = NULL;

while(tmp->next != NULL){

tmp = tmp->next;

}

tmp->next = newnode;

}

}

void menu(){

clrscr();

printf("\n1. Add new patient to the end");

printf("\n2. Add new patient to the beginning");

printf("\n3. Move patient to front");

printf("\n4. Print out the entire list");

printf("\n5. Remove a node from the list");


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printf("\n6. Quit the program\n\n");

}

void display(){

struct list *temp;

temp = head;

if(temp == NULL){

printf("List empty");

}

while(temp != NULL){

printf("\t\n%d\n", temp->id);

printf("Name: %s\n", temp->name);

printf("Critical State: %d\n", temp->cstate);

printf("Ailement: %s\n", temp->ail);

temp = temp->next;

}

getch();


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}

void removenode(struct list *head){

int id;

int found=0;

struct list *current = head;

printf("Enter ID: ");

fflush(stdin);

scanf("%d", &id);

while(current->next != NULL) {

if(current->id == id){

found++;

break;

}

current = current->next;

}

if(found == 0){

printf("No patient found for that ID");

}


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free(current);

getch();

}

void preempt(){

struct list *tmp, *current;

int id;

tmp = head;

printf("\nEnter patient ID :");

scanf("%d", &id);

while(tmp->id != id){

current = tmp;

tmp = tmp->next;

if(tmp == NULL){

printf("End of list");

break;

}

}

current->next = tmp->next;

tmp->next = head;


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head = tmp;

//printf("%s", tmp->name);

getch();

}


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Queues

/*

ABC CAR PARK SYSTEM v1.0

ORGINALLY CODED BY - Praveena Sarathchandra on 17th

September 2009

2009

*/

#include

#include

#include

#include

#include

#include

#include

#define MAX 10

struct cars{

int plateno;

time_t intime;

int moved;

char vtype;

};


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struct queue{

int front;

int rear;

int count;

struct cars newq[MAX];

}mainq;

//**************FUNCTION PROTOTYPES*******************\\

void header();

void menu();

void showInput(struct queue *mainq);

void input(struct queue *mainq, int plate, char type, time_t intime, char vtype);

void init(struct queue *mainq);

void display(struct queue *mainq);

void printReceipt(struct queue *mainq, char type, int plate, time_t inTime, int slot, char

vtype);

void fullMsg();

void emptyMsg();

main(){

int choice;

init(&mainq);


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do{

menu();

scanf("%d", &choice);

switch(choice){

case 1:

showInput(&mainq);

break;

case 2:

display(&mainq);

break;

}

}while(choice != 3);

if(choice == 3)

exit(1);

}

//**************FUNCTION DECLARATION*******************\\


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void menu(){

clrscr();

header();

printf("\n\t\t1. Input Vehicle info\n");

printf("\t\t2. Show Stats\n");

printf("\t\t3. Exit\n\n");

printf("\n\n\t\tEnter Choice : ");

}

void header(){

printf("\n\t\t************************************************* ");

printf("\n\t\t------------- ABC VEHICLE PARK -----------------\n");

printf("\t\t*************************************************\ n\n");

}

void showInput(struct queue *mainq){

int plate;

char type;

char vtype;

time_t intime;

clrscr();

header();


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printf("\n\t\tArrival(a) / Departure(d): ");

fflush(stdin);

scanf("%c", &type);

type = tolower(type);

switch(type){

case 'a':

if(mainq->count >=10){

fullMsg();

} else {

clrscr();

header();

printf("\t\t\t [Enter Vehicle details]\n\n");

printf("\t\tEnter vehicle type (Car - c, Van - v):");

fflush(stdin);

scanf("%c", &vtype);

vtype = tolower(vtype);

printf("\t\tLicense Plate No:");

fflush(stdin);


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scanf("%d", &plate);

intime = time(NULL);

input(mainq, plate, 'a', intime, vtype);

}

break;

case 'd':

time_t timenow;

clrscr();

header();

printf("\n\t\t\t[Enter Vehicle details]\n\n");

printf("\t\t\tLicense Plate No:");

fflush(stdin);

scanf("%d", &plate);

timenow = time(NULL);

input(mainq, plate, 'd', timenow,NULL );

break;


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}

}

void init(struct queue *mainq){

mainq->front = 0;

mainq->rear = -1;

mainq->count = 0;

for(int i=0;inewq[i].plateno = 0;

mainq->newq[i].intime = 0;

mainq->newq[i].moved = 0;

}

}

void input(struct queue *mainq, int plate, char type, time_t intime, char vtype){

int i=0,j=0;

time_t outTime;


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switch(type){

case 'a':

mainq->newq[++mainq->rear].plateno = plate;

mainq->newq[mainq->rear].intime = intime;

mainq->newq[mainq->rear].vtype = vtype;;

mainq->count++;

printReceipt(mainq, 'a', plate, intime, mainq->rear+1, vtype);

getch();

break;

case 'd':

i=0;

//get the index for the given plate #

while(mainq->newq[i].plateno != plate){

i++;

}

outTime = time(NULL);


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if(i == 0){

for(j=i+1; jrear ; j++){

mainq->newq[j-1].plateno = mainq-

>newq[j].plateno;

mainq->newq[j-1].moved = mainq-

>newq[j].moved;

mainq->newq[j-1].moved +=1;

mainq->newq[j-1].intime = mainq-

>newq[j].intime;

}

mainq->rear--;

} else if( (i > 0) && (i < mainq->rear)) {

for(j=0;jnewq[j].moved++;

}

for(j=i+1; jrear ; j++){


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mainq->newq[j-1].plateno = mainq->newq[j].plateno;

mainq->newq[j-1].moved = mainq->newq[j].moved;

mainq->newq[j-1].moved +=1;

mainq->newq[j-1].intime = mainq->newq[j].intime;

mainq->newq[j].plateno = 0;

mainq->newq[j].moved= 0;

mainq->newq[j].intime=0;

}

mainq->rear--;

}else if( i == mainq->rear){

mainq->newq[i].plateno = 0;

mainq->newq[i].moved= 0;

mainq->newq[i].intime=0;

mainq->rear--;

}

mainq->count--;

printReceipt(mainq, 'd', plate, outTime, mainq->rear+1, mainq-

>newq[i].vtype);

getch();


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break;

}

}

void display(struct queue *mainq){

if(mainq->count == 0){

emptyMsg();

} else {

clrscr();

header();

for(int i=0;irear;i++){

printf("\n\t\t-------SLOT %d--------\nLICENSE PLATE #%d \nVehicle

type: %c \nTIME IN: %s \nMOVED TIMES: %d\n", i, mainq->newq[i].plateno, mainq-

>newq[i].vtype, asctime(localtime(&mainq->newq[i].intime)), mainq->newq[i].moved);

}

getch();

}

}

void printReceipt(struct queue *mainq, char type, int plate, time_t inTime, int slot, char

vtype){

time_t timeNow;

timeNow = time(NULL);


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float charge=0;

float duration;

float hrs;

float discount=0;

float total;

header();

switch(type){

case 'a':

clrscr();

header();

printf("\n\t\t\t ------------------------");

printf("\n\t\t\t| New Car Added to Queue |");

printf("\n\t\t\t ------------------------");

printf("\n\t\tLICENSE PLATE : %d\n", plate);

printf("\n\t\tVehicle Type: %c\n", vtype);

printf("\n\t\tTime in: %s\n", asctime(localtime(&inTime)));

printf("\t\tSlot : #%d\n", slot);

break;


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case 'd':

duration = difftime(inTime,mainq->newq[slot].intime);

switch(vtype){

case 'c':

if( duration


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}

if(mainq->newq[slot].moved > 10){

discount = charge * (1/100);

}

total = charge - discount;

clrscr();

header();

printf("\n\t\tLICENSE PLATE : %d\n", plate);

printf("\t\tSlot : #%d\n", slot);

printf("\n\t\tTime in: %s", asctime(localtime(&mainq->newq[slot].intime)));

printf("\n\t\tTime out: %s", asctime(localtime(&inTime)));

printf("\n\t\tDuration: %f", duration);

printf("\n\t\tCharge: %.2f", charge);

printf("\n\t\tTimes moved: %d", mainq->newq[slot].moved);

printf("\n\t\tDiscount %.2f:", discount);

printf("\n\t\t_________________________");

printf("\n\t\tTOTAL CHARGE: %.2f", total);

printf("\n\t\t_________________________");

break;

}


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}

void fullMsg(){

printf("\n\t[THE QUEUE ISFULL]\n");

getch();

}

void emptyMsg(){

printf("\n\t\t\t[THE QUEUE IS EMPTY]\n");

getch();

}


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Bibliography


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WorkLoad Metrix

Task Praveena Pulasthi

Preparation and planning

Analyze the requirements of system

Design

Propose a solution

Research and select appropriate methodology for functions

Circular Queues implementation

Linked List implementation

Development

Develop necessary modules of the proposed

system

Circular Queues implementation

Linked List implementation

Testing

Testing and Debugging

Documentation and Presentation

Report and documentation

Presentation


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ASIA PACIFIC INSTITUTE OF INFORMATION TECHNOLOGY

AAPP001-3-2-DATABASES & DATA STRUCTURES

Assignment Marking Scheme

Group:

Student Id:

Criteria

Marks

Comments

Implementation (12 marks)

y Program logic (3)

y Validation checks (3)

y Expected results (3)

y Workable system (3)

Efficiency memory utilisation (8 marks)

y Efficient memory utilisation (4)

y Modular/ Compact program (3)

y No user ofunnecessary global

vars, go to statements etc (2)

Comprehension (12 marks)

y Explanation ofcode (4)

y Capability to debug errors (4)

y Logic justification (4)

Assignment contribution (8 marks)

y Workload distribution (6)

y Initiative taken (2)

Total: 40 marks


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ASIA PACIFIC INSTITUTE OF INFORMATION TECHNOLOGY

AAPP001-3-2-DATABASES & DATA STRUCTURES

Assignment Marking Scheme

Group:

Student Id:

Criteria

Marks

Comments

Implementation (12 marks)

y Program logic (3)

y Validation checks (3)

y Expected results (3)

y Workable system (3)

Efficiency memory utilisation (8 marks)

y Efficient memory utilisation (4)

y Modular/ Compact program (3)

y No user ofunnecessary global

vars, go to statements etc (2)

Comprehension (12 marks)

y Explanation ofcode (4)

y Capability to debug errors (4)

y Logic justification (4)

Assignment contribution (8 marks)

y Workload distribution (6)

y Initiative taken (2)

Total: 40 marks


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Databases and Data Structures [Individual Assignment]

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