today s agenda
Download
Skip this Video
Download Presentation
Today’s Agenda

Loading in 2 Seconds...

play fullscreen
1 / 13

Today’s Agenda - PowerPoint PPT Presentation


  • 65 Views
  • Uploaded on

Today’s Agenda. Sorting. Insertion Sort – Top Down Design. Sort list: A[0] to A[N-1]. Sort list: A[0] to A[N-2]. Insert A[N-1] into sorted list. Sort list: A[0] to A[N-3]. Insert A[N-2] into sorted list. …. Sort list: A[0]. Insert A[1] into sorted list.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' Today’s Agenda' - zenaida-stephenson


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
insertion sort top down design
Insertion Sort – Top Down Design

Sort list: A[0] to A[N-1]

Sort list: A[0] to A[N-2]

Insert A[N-1] into sorted list

Sort list: A[0] to A[N-3]

Insert A[N-2] into sorted list

Sort list: A[0]

Insert A[1] into sorted list

slide3

int insert(int res[],int x,int size) {

int i=0, j, k, flag=0;

if(size == 0)

{

res[0] = x; size = size + 1;

}

else

{

while(i< size)

{

if(res[i] < x || res[i] == x)

i++;

else {

j = i; flag = 1; break;

}

}

if(flag == 1)

{

for(k=size;k>j;k--)

res[k] = res[k-1];

res[k] = x;

}

else

res[i] = x;

size = size + 1;

}

return size; }

slide4
#define MAX 20

int main()

{

int a[10] = {50,16,20,35,18,8,14,41,3,39};

int res[MAX];

int i,size=0;

size = insertInOrder(res,a[i],size);

for(i=0;i<size;i++)

printf("%d ",res[i]);

printf("\n");

}

slide6
Consider a police department containing N police officers. Each police officer contains the following information in his record.

1. An unsigned int id representing the unique identification number of the police.

2. A character array of maximum 30 characters for the name.

3. An unsigned int sc for the number of cases solved

4. An unsigned int ac for the number of cases attended

5. An unsigned int age for storing the age of the police

6. An unsigned int jyr to store the joining year.

  • A police department is a list containing maximum of MAX_POL number of police officers where MAX_POL is 1000.
slide7
readPoliceDept()
  • rank()
  • countPoliceOnRank()
  • countPoliceService()
  • sortPoliceDept()
  • comparePolice()
  • printPoliceDept()
operations
Operations

(1) rank()

  • The rank of police officer is computed as follows.
  • rank = (100 * solvedCases / attendedCases)
  • if attendedCases = 0, rank = 0;
slide9
comparePolice()
    • compares police p1 and p2 based on rank, if rank is same then compare based on id .
slide10

PoliceDef.h

#define NAME_LEN 50 // Maximum length of Name

#define MAX_POL 1024 // Max number of Police in one Dept

typedef unsigned int ID; // Unique Police identification number

typedef char NAME[NAME_LEN]; // Name of Police

typedef unsigned int solvedCases; // Number of cases solved

typedef unsigned int attendedCases; // Total no. of cases attended

typedef unsigned int AGE; // Police officer\'s Age

typedef unsigned int joiningYear; // Police officer\'s joining year

typedef unsigned int noOfPolice; // number of Police officers in Dept

policedef h
PoliceDef.h

typedef enum { GREATER = 1, EQUAL = 0, LESSER = -1 } ORDER;

typedef struct {

ID id;

NAME name;

solvedCases sc;

attendedCases ac;

AGE age;

joiningYear jyr;

} police; // The police structure

typedef police policeDept[MAX_POL]; // the policeDept

policeops h
PoliceOps.h
  • Extern void readPoliceDept(policeDeptpdt, noOfPolice N);
  • Extern float rank(police p);
  • Extern int countPoliceOnRank (policeDept pdt, noOfPolice N);
    • Count the number of police who has rank more 75.
  • Extern int countPoliceService(policeDept pdt, noOfPolice N);
    • Count the number of police who has completed 25 years of service.
  • Extern order comparePolice(policep1,police p2);
  • Extern void sortPoliceDept(policeDept pdt, noOfPolice N);
    • Sort police based on their ranks.
  • Extern void printPoliceDept(policeDept pdt, noOfPolice N);
comparepolice
comparePolice()

ORDER comparePolice (police p1, police p2)

{

if(rank(p1)<rank(p2))

return LESSER;

else if(rank(p1) == rank(p2))

{

if(p1.id < p2.id)

return LESSER;

else if(p1.id == p2.id)

return EQUAL;

else

return GREATER;

}

else

return GREATER;

}

ad