Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 9 pdf
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 9 pdf
... + E) to compute G
.
•
O(VE) to run BELLMAN-FORD.
•
(E) to compute w.
•
O(V
2
lg V +VE) to run Dijkstra’s algorithm
|
V
|
times (using Fibonacci heap).
•
(V
2
) to compute D matrix.
Total: ... t
(k−1)
kj
.
T
RANSITIVE-CLOSURE(E, n)
for i ← 1 to n
do for j ← 1 to n
do if i = j or (i, j) ∈ E[G]
then t
(0)
ij
← 1
else t
(0)
ij
← 0
for k ← 1 to n
do for i ← 1 to n
do for j...
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 1 pptx
... Data Structures
Lecture Notes 14-1
Solutions 14 -9
Preface
This document is an instructor’s manual to accompany Introduction to Algorithms,
Second Edition, by Thomas H. Cormen, Charles E. Leiserson, ... invariants is like mathematical induction:
Instructor’s Manual
by Thomas H. Cormen
Clara Lee
Erica Lin
to Accompany
Introduction to Algorithms
Second Edition
by...
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 2 pps
... 5.
•
MAX-HEAPIFY is applied to subtrees rooted at nodes (in order): 16, 2, 3, 1, 4.
1
23
4567
891 0
1
23
4567
891 0
4
13
291 0
14 8 7
16
41 23 16 9 10 14 8 7
16
14 10
893
241
7
A
i
234567 891 01
Correctness
Loop ... E
2
[
X
j
]
=
0
2
·
99
100
+
100
2
·
1
100
− 1
2
= 100 − 1
= 99 .
Summing up the variances of the X
j
gives Var
[
V
n
]
= 99 n.
5-12 Solutions for Chapter 5...
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 3 docx
... 7-3
81640 395
p,j ri
81640 395
prj
18640 395
p,i rj
18640 395
p,i rj
1864 0 395
prji
18640 395
prji
13640 895
prji
13640 895
pri
16540 893
pri
i
A[r]: pivot
A[j r–1]: not yet examined
A[i+1 j–1]: known to ... overview
[The treatment in the second edition differs from that of the Þrst edition. We use
a different partitioning method—known as “Lomuto partitioning”—in the second...
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 4 pot
... pointer) to the new slot, and updating the pointer in the slot
that pointed to j to point to the new slot. Then insert the new element in
the now-empty slot as usual.
To update the pointer to j , ... lists. If we want to delete elements, it’s better to
use doubly linked lists.]
•
Slot j contains a pointer to the head of the list of all stored elements that hash
to j
[or t...
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 5 potx
... O(h) time,
analogous to the changes we made for persistence in insertion. But to do so
without using parent pointers we need to walk down the tree to the node to be
deleted, to build up a stack ... keys are not distinct, because in order to Þnd the
path to the node to delete—a particular node with a given key—we have to
make some changes to how we store things in the tre...
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 6 pps
... row-major order, i.e.,
row-by-row from top to bottom, and left to right within each row. Column-
major order (column-by-column from left to right, and top to bottom within
each column) would also ... long it takes to get through S
1,1
.
•
If j ≥ 2, have two choices of how to get to S
1, j
:
•
Through S
1, j −1
, then directly to S
1, j
.
•
Through S
2, j −1
, then transfer ove...
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 7 ppsx
... to go to get to 1 or 1/4, starting from 1/2, rate
of increase of differs.
•
For α to go from 1/ 2to1 ,num increases from size /2tosize, for a total
increase of size /2. increases from 0 to size. ... sorting A and B into monotoni-
cally increasing order works as well.
Lecture Notes for Chapter 16: Greedy Algorithms 16-3
Solution to S
ij
is (solution to S
ik
) ∪
{
a
k
}
∪ (s...
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 8 potx
... edge it visits to the cycle C, which is then added to the Euler
tour T , when we return from a call to VISIT(u), all edges entering or leaving
vertex u have been added to the tour. When we advance ... B to the end of A, instead
splice B into A right after the Þrst element of A. We have to traverse B to update
representative pointers anyway, so we can just make the last element o...
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 10 potx
... the input to a half-cleaner is a bitonic 0-1 sequence, then for the output:
•
both the top and bottom half are bitonic,
•
every element in the top half is ≤ every element in the bottom half, ... 27-5
Bitonic sorter:
≤
bitonic
sorter
half-cleaner
n
n/2
n/2
n/2
n/2
sorted
bitonic
bitonic
sorted
sorted
bitonic
sorter
0
0
0
0
1
0
1
1
0
0
0
0
1
0
1
1
0
0
1
1
1
0
0
0
0
0
0
0
0
1
1
1
bitonic ... a...
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