w11 Priority Queue

5/28/2018 w11 Priority Queue

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Bi 8: Hngutin

Gingvin: Hong Th ip

Khoa Cng nghThng tin HCng Ngh


2/42

Nidung

KDLTT hng u tin

Ccphngphp ci t

ng dng: xy dng m Huffman

INT2203


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KDLTT hngutin(priority queue)

L tp hp trong mi phnt l mt cp(gi tr utin,i tng) ta c thso snhccc

gi tr utin

Cc php ton insert(k, o) xen vo hngu

tini tngo c gi tr utin k.

findMin() tmi tngc gi

tr utin nhnht.Thchincnuhng khng rng

removeMin() loibv trvi tngc gi tr utinnhnht. Thchin cnuhng khng rng.

findMinKey() tm gi tr utin

nhnht.Thchin cnuhng khng rng

size()

isEmpty()

ngdng

Qunlbngthng Sdngtrong thitkcc

thutton (Huffman)

INT2203


4/42

Minh ha

INT2203

Php ton Output Hng utin

insert(5,A) - {(5,A)}

insert(9,C) - {(5,A), (9,C)}

insert(3,B) - {(3,B), (5,A), (9,C)}

insert(7,D) - {(3,B), (5,A), (7,D), (9,C)}

findMin() B {(3,B), (5,A), (7,D), (9,C)}findMinKey() 3 {(3,B), (5,A), (7,D), (9,C)}

removeMin() - {(5,A), (7,D), (9,C)}

size() 3 {(5,A), (7,D), (9,C)}

findMin () A {(5,A), (7,D), (9,C)}

removeMin() - {(7,D), (9,C)}removeMin() - {(9,C)}

removeMin() - {}

removeMin() error {}

isEmpty() true {}


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Nidung

KDLTT hng u tin

Ccphngphp ci t


INT2203


6/42

Ccphngphp ci t

INT2203

findMin insert removeMin

Mng csp ? ? ?

Mngkhng sp ? ? ?

DSLKcsp ? ? ?

DSLK khng sp ? ? ?

Cy TKNP ? ? ?

Cy thtbphn(heap)

? ? ?


7/42

Ci thngutin bidanh sch

V d Minh hamng

Danh schcsp

Q = {(3,B), (5,A), (7,D), (9,C)}

findMin()

insert(8, E)removeMin()

Danh schkhng sp

Q = {(7,D), (3,B), (9,C), (5,A)}

findMin()

insert(8, E)

removeMin()

INT2203

9,C 7,D 5,A 3,B

9,C 7,D 5,A 3,B

9,C 8,E 7,D 5,A 3,B

9,C 8,E 7,D 5,A

7,D 3,B 9,C 5,A

7,D 3,B 9,C 5,A

7,D 3,B 9,C 5,A 8,E

7,D 9,C 5,A 8,E


8/42

Ci t hngutin bicy tm kim nh phn

INT2203

7,D

9,C3,B

5,A2,F 8,E

7,D

9,C3,B

5,A2,F 8,E

7,D

9,C3,B

5,A2,F 8,E

4,G

7,D

9,C3,B

5,A2,F 8,E

4,G

1. hngutin 2. findMin()

3. insert(4,G)4. removeMin()


9/42

Cy thtbphn(heap)

Cy nhphn hon ton c ttccc mccacyu

khng thiu nhno, trmcthpnht clp yktbn tri

Min heap l mtcy nhphnhon ton vitnh cht kha camt nhbtknh

hnhocbngkha caccnhcon can

Max heap Citheap

c thdng cutrc lin kt(con tr)

c thdng mng

cao caheap l log(n) INT2203

2

35

79 8

0

1 2

3 4 5

2 5 3 9 7 8


10/42

Ci thngutin biheap

findMin?

insert?

removeMin?

INT2203

2,F

3,B5,A

7,D9,C 8,E

0

1 2

3 4 5

2,F 5,A 3,B 9,C 7,D 8,E


11/42

Xen thm vo heap

Phpinsertcahngutintng ngviphpxen thm mtphntckhakvo heap

Thutton Tm tivtrz aphntmivo (lnhcuimi)

Luphntc khakvoz

Khi phctnh chtthtbphncaheap

INT2203

2

65

79

insertion node

2

65

79 1

z

z

z

z


12/42

Upheap

Sau khi xen thm mtkhak mi, heap c thmt itnh chtthtbphn

Thutton upheap khi phclitnh chtny bngcchochkdctheong it nhmi hngtigc

Upheap dnglikhiktintigchocmt nhc kha cacha k

V heap ccaoO(logn), upheap thchintrong thigianO(logn)

INT2203

2

15

79 6z

1

25

79 6z


13/42

Loimtphntkhiheap

Php removeMin cahngutintng ngviphp loigccaheap

Thutton Thay thgcbng nhcuicngw

Giiphngnhw Khi phctnh chttht

bphncaheap

INT2203

2

65

79

nhcui

w

7

65

9

w


14/42

Downheap

Sau khi thay th nhgcvi nhcui(c khak), heap c thmtitnh chtthtbphn

Thutton downheap khi phclitnh chtny bngcchochnhc khakdctheong itgcxung

Upheap dngkhiktintimtl hay mt nhc cc kha con k Do heap ccaoO(logn), downheap thchintrong thigianO(logn)

INT2203

7

65

9

w

5

67

9

w


15/42

Cpnhtcon trti nhcui

Dng trong citheap bngcutrc lin kt

C thtm chchonhsthm vo bngcchitheo hnh trnhgmO(logn) nh iln tikhi gpmtcon tri hocgpgc

Nugpmtcon tri th chuynsang con phi ixungtikhi gpmtl

p dngthuttontngtcho cpnhtcon trti nhcuitrongphp loibmt nh

INT2203


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Minh hacithngutin

INT2203

(4,C)

(5,A)

(15,K)

(16,X) (25,J)

(9,F)

(14,E) (12,H)

(6,Z)

(7,Q) (20,B)

(11,S) (18,W)

heap

last

comp


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(18,W)

insert(2,T)

INT2203

(4,C)

(5,A)

(15,K)

(16,X) (25,J)

(9,F)

(14,E) (12,H)

(6,Z)

(7,Q) (20,B)

(11,S)


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(18,W)

insert(2,T)

INT2203

(4,C)

(5,A)

(15,K)

(16,X) (25,J)

(9,F)

(14,E) (12,H)

(6,Z)

(7,Q) (20,B)

(11,S) (2,T)


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(18,W)

insert(2,T)

INT2203

(4,C)

(5,A)

(15,K)

(16,X) (25,J)

(9,F)

(14,E) (12,H)

(6,Z)

(7,Q) (20,B)

(11,S) (2,T)

o(2,T) v(20,B)


20/42

(6,Z)

(20,B)(18,W)

insert(2,T)

INT2203

(4,C)

(5,A)

(15,K)

(16,X) (25,J)

(9,F)

(14,E) (12,H)

(7,Q) (2,T)

(11,S)

o(2,T) v(6,Z)


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(2,T)

(20,B)(18,W)

insert(2,T)

INT2203

(4,C)

(5,A)

(15,K)

(16,X) (25,J)

(9,F)

(14,E) (12,H)

(7,Q) (6,Z)

(11,S)

o

(2,T) v(4,C)


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(4,C)

(20,B)(18,W)

insert(2,T)

INT2203

(2,T)

(5,A)

(15,K)

(16,X) (25,J)

(9,F)

(14,E) (12,H)

(7,Q) (6,Z)

(11,S)

Sau thigian O (log n) th cy lithnh mtheap


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(18,W)

removeMin()

INT2203

(4,C)

(5,A)

(15,K)

(16,X) (25,J)

(9,F)

(14,E) (12,H)

(6,Z)

(7,Q) (20,B)

(11,S)


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(18,W)

removeMin()

INT2203

(4,C)

(5,A)

(15,K)

(16,X) (25,J)

(9,F)

(14,E) (12,H)

(6,Z)

(7,Q) (20,B)

(11,S)


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removeMin()

INT2203

(18,W)

(5,A)

(15,K)

(16,X) (25,J)

(9,F)

(14,E) (12,H)

(6,Z)

(7,Q) (20,B)

(11,S)

o(18,W)v (5,A)


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removeMin()

INT2203

(5,A)

(15,K)

(16,X) (25,J)

(9,F)

(14,E) (12,H)

(6,Z)

(7,Q) (20,B)

(11,S)

o(18,W)v (9,F)

(18,W)


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removeMin()

INT2203

(5,A)

(15,K)

(16,X) (25,J)

(9,F)

(14,E) (12,H)

(6,Z)

(7,Q) (20,B)

(11,S)

o(18,W)v (12,H)

(18,W)


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removeMin()

INT2203

(5,A)

(15,K)

(16,X) (25,J)

(9,F)

(14,E)

(12,H)

(6,Z)

(7,Q) (20,B)

(11,S)(18,W)

nhcui


29/42

phctp

phctpkhng gian Ci bngcutrc lin kt(dng con tr): O(n)

Ci bngcutrc vector (mng): tlviN (ccamng)

phctpthigian

INT2203

Php ton Thigian

size, isEmpty O (1)

findMin, findMinKey O (1)

insert O (log n)

removeMin O (log n)


30/42

Tngkt

INT2203

findMin insert removeMin

Mng csp O(1) O(n) O(1)

Mngkhng sp O(n) O(1) O(n)

DSLKcsp O(1) O(n) O(1)

DSLK khng sp O(n) O(1) O(n)

Cy TKNP O(h) O(h) O(h)

Cy thtbphn(heap)

O(1) O(logn) O(logn)


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Nidung

KDLTT hng u tin

Ccphngphp ci t


INT2203


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Nn dliu

Giscnnn mttpdliucha100000 k ttbng6 chci (ta nf).

Mdi gingnhau (a) biudinmichci bi3 bit

(thay v 8 bit nh thngl) tlnn = 3/8

Mdi khc nhau (b) dng khi ta bittnsutcacc

chci

gn m ngnnhtcho chci

xuthinnhiunht kchthcfile nn:

(451 + 133 + 123 + 163 + 94 +54) 1000 = 224 000 bits

tlnn = 0.28

Chci a b c d e f Tm 000 001 010 011 100 101

INT2203

Chci a b c d e f Tnsut(K) 45 13 12 16 9 5

Tm 0 101 100 111 1101 1100

(a)

(b)

Ch : khng c m noclmtintcam khc

o gil m tint(prefix code)

o mc ch: phcvgiinn

Bi ton: xy dngm tintvitlnn thpnht

o Ligii: m Huffman


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M Huffman

Biudinm tint didngcy nhphn timi nh, nhnh tric

gnnhn l 0, nhnh bn phicgnnhn l 1

mik t c lutrong mtnhl tm camik tl xu bit

tothnh tcc nhn trnng itgcti nhlchak t

Thutton Huffman sdnghngutinxy dngmtint didngcy nhphn M sinh ra gilm Huffman

INT2203

Chci a b c d e f Tnsut(K) 45 13 12 16 9 5

Tm 0 101 100 111 1101 1100

100

45(a)

55

25 30

12(c)

13(b)

14 16(d)

5(f) 9(e)

0

0

0 0

0

1

1

1

1

1


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Thutton Huffman

Vimik txuthintrong xungun, ta tora mt nhchakt gnvigi tr utin bngtn

sut

Ttpcc cy chc mt nh, timi bcta kthphai cy thnhmtcy nhcha sgnvigi tr utin

bngtng utin cc con

ta cnchnhai cy nhphn c

mc utin nhnhtkthpthnh mt dng hngutin.

INT2203

Algorithm HuffmanCoding(S,F)Input: Bngchci S v bngcc tn

sutFOutput: cy HuffmanTo ra mt nh cho mik ttrong SKhi to hng u tin P cha ccnh

nyfori=0to n-1do

v1 P.removeMin()v2 P.removeMin()Toranhv mivi

leftChild(v) v1

rightChild(v) v2f(v) f(v1) + f(v2)

P.insert(v)


35/42

INT2203

45(a)

12(c)

13(b)

16(d)

5(f)

9(e)


36/42

INT2203

45(a)

12(c)

13(b)

14 16(d)

5(f)

9(e)

0 1


37/42

INT2203

45(a)

25

12(c)

13(b)

14 16(d)

5(f)

9(e)

0

0 1

1


38/42

INT2203

45(a)

25 30

12(c)

13(b)

14 16(d)

5(f)

9(e)

0 0

0

1

1

1


39/42

INT2203

45(a)

55

25 30

12(c)

13(b)

14 16(d)

5(f)

9(e)

0

0 0

0

1

1

1

1


40/42

INT2203

100

45(a)

55

25 30

12(c)

13(b)

14 16(d)

5(f)

9(e)

0

0

0 0

0

1

1

1

1

1


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Nidung

KDLTT hng u tin

Ccphngphp ci t


INT2203


42/42

Chunbbi ti

c chng 16(Thitkthutton)

Date post:	18-Oct-2015
Category:	Documents
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