Becoming a Good Problems Solver
― Challenging ICPC-2003/2004
Yaw-Ling Lin
Dept Computer Science & Info. Management,
Providence University, Taichung, Taiwan.
1
27th Annual ACM ICPC World Finals
•
•
•
•
•
•
•
68 final teams
3,850 teams
1,329 universities
68 countries
106 sites
Champions: Warsaw University.
The 27th Annual ACM International Collegiate
Programming Contest, March 25, 2003, in Beverly Hills,
California.
2
ICPC2002 Kaohsiung: 8 Problems
3
ICPC Kaohsuing, Nov 09, 2002
Rank
Team
1
Zhongshan University=>Bearcat (6)
2
National University of Singapore=>NUS-SOC (6)
3
National Chiao-Tung University=>NCTU Fresh Team (6)
4
Chiao Tung University=>Invulnerable Champion (6)
5
The Chinese University of Hong Kong=>CUHK Libra (5)
6
National Tsing Hua University=>Eagle (5)
7
National Taiwan University=>Pheonix (5)
8
The University of Hong Kong=>HKU CE (5)
9
National Taiwan University=>Yatta (5)
10
National Taiwan University=>Joking Kids (5)
4
ICPC2003 World: 10 Problems
•
•
•
•
•
A: Building Bridges
B: Light Bulbs
C: Riding the Bus
D: Eurodiffusion
E: Covering Whole
Holes
•
•
•
•
•
F: Combining Images
G: A Linking Loader
H: A Spy in the Metro
I: The Solar System
J: Toll
5
ICPC 2003 Results
ICPC2003 (前55名)
9
15
13
13
13
7
9
10
6
5
5
1
5
1
0
9
8
8
7
6
5
4
3
4
3
6
2003 ACM-ICPC World Finals
Rank
Name
Solved
1
Warsaw University (1) – Poland
9
2
Moscow State University (1) – Russia
8
3
St Petersburg Institute (5) – Russia
7
4
Comenius University – Slovakia
7
5
Tsinghua University – China
7
6
Shanghai JiaoTong University – China
7
7
Saratov State University – Russia
7
8
Zhongshan University (13) – China
6
13
California Inst Tech, UC Berkeley – USA
6
13
Fudan; Zhejiang University – China
6
13
U Tokyo, U Singapore – Japan, Singapore
6
21
Carnegie Mellon University (9)
5
30
Harvard (13)
4
43
National Chiao-Tung University (13)
3
7
ICPC Problem Sets
• ICPC2003-World
– http://icpc.baylor.edu/icpc/Finals/Problems.html
• ICPC2002-Kaohsiung
– http://icpc.baylor.edu/past/icpc2003/Regionals/Rep
orts/icpc.baylor.edu/icpc/regionals/Kaohsiun02/ind
ex.html
8
Various Kinds of Problems
•
•
•
•
•
•
Algorithms
Simulation
System Programming
Geometry
Combinatorics
Modeling, Situation Analysis
9
G. Polya, "How to Solve It“, 1957
1. Understanding The
Problem
2. Devising A Plan
3. Carrying Out The
Plan
4. Looking Back
- If you can't solve a problem, then there is an easier
problem you can solve: find it.
10
UNDERSTANDING THE
PROBLEM
• First. You have to understand the problem.
• What is the unknown? What are the data? What is the
condition?
• Is it possible to satisfy the condition? Is the condition
sufficient to determine the unknown? Or is it
insufficient? Or redundant? Or contradictory?
• Draw a figure. Introduce suitable notation.
• Separate the various parts of the condition. Can you
write them down?
11
DEVISING A PLAN
• Second. Find the connection between the data and the
unknown. Consider auxiliary problems if an immediate
connection cannot be found.
• Have you seen it before?
• Do you know a related problem?
• Look at the unknown! Think of a familiar problem having the
same or a similar unknown.
• Could you restate the problem? Could you restate it differently?
Go back to definitions.
• If you cannot solve the proposed problem try to solve first
some related problem.
• Did you use all the data? Did you use the whole condition?
12
CARRYING OUT THE PLAN
• Third. Carry out your plan.
• Carrying out your plan of the solution, check
each step. Can you see clearly that the step is
correct? Can you prove that it is correct?
13
LOOKING BACK
• Fourth. Examine the solution obtained.
• Can you check the result? Can you check the
argument?
• Can you derive the solution differently? Can
you see it at a glance?
• Can you use the result, or the method, for
some other problem?
14
Look at the Problems
-- Problems Types
•
•
•
•
•
•
•
Algorithms
Dynamic Programming
Simulation
System Programming
Geometry
Combinatorics
Problem Solving
15
Basic Techniques
• English / Problem Modeling
• Combinatorics / Discrete Math
– Counting
– Recurrence
– Graphs, Trees, Connectivity
• Algorithm:
– Dynamic Programming
– Graph Searching: DFS/BFS, Shortest Path, MST, Tree
Traversal
• Efficient (Human) Programming
16
Training
•
•
•
•
•
•
•
•
•
Solve as many past ICPC problems as possible.
English reading and Problem Modeling  read Polya.
Thoroughly understand discrete math / structure
Understand / practice algorithmic techniques: read algorithm +
data structure text book
Optional: Computational Geometry, String algorithms
Programming efficiency: routine gadgets, fast-typing,
debugging tricks, writes, writes. writes …
Communications / discussions between members.
Leadership / role-playing / execution.
Time devotion.
17
Members
•
•
•
•
English / Problem Understanding
Problem Modeler / Dispatcher
Human-Time Efficient Programmers
Team Work
18
ICPC2003 World: 10 Problems
•
•
•
•
•
A: Building Bridges
B: Light Bulbs
C: Riding the Bus
D: Eurodiffusion
E: Covering Whole
Holes
•
•
•
•
•
F: Combining Images
G: A Linking Loader
H: A Spy in the Metro
I: The Solar System
J: Toll
19
A: Building Bridges
City 1
City 3
No bridges are needed
City 1
with bridges
City 2
No bridges are possible
City 4
City 4
with bridges
20
A: Building Bridges
• Bridges may be built only on the grid lines that form the edges
of the squares. Each bridge must be built in a straight line and
must connect exactly two buildings.
• For a given set of buildings, you must find the minimum
number of bridges needed to connect all the buildings. If this is
impossible, find a solution that minimizes the number of
disconnected groups of buildings.
• Among possible solutions with the same number of bridges,
choose the one that minimizes the sum of the lengths of the
bridges, measured in multiples of the grid size. Two bridges
may cross, but in this case they are considered to be on
separate levels and do not provide a connection from one
bridge to the other.
21
A: Building Bridges
22
Building Bridges: Solutions
City 1
City 3
No bridges are needed
City 1
with bridges
City 2
No bridges are possible
City 4
City 4
with bridges
• Block: Nodes
• Connectivity: Weighted Edges
• Model: Minimum Spanning Tree
23
B: Light Bulbs
first switch
first bulb
second switch
third switch
second bulb
third bulb
nth switch
(n-1)st bulb
nth bulb
24
B: Light Bulbs
• Each remaining switch (1 < i < n) toggles the states of the
three bulbs with indices i – 1, i, and i + 1.
• The minimum cost of changing a row of bulbs from an initial
configuration to a final configuration is the minimum number
of switches that must be flipped to achieve the change.
• 01100 represents a row of five bulbs in which the second and
third bulbs are both ON. You could transform this state into
10000 by flipping switches 1, 4, and 5, but it would be less
costly to simply flip switch 2.
• Decimal integers are used instead of binary for the bulb
configurations. Thus, 01100 and 10000 are represented by the
decimal integers 12 and 16.
25
B: Light Bulbs
26
Light Bulbs: Solution
• 10100  01110 == 00000  11010
• x1 xor x2 = b1
• x1 xor x2 xor x3 = b2
 x3 = b1 xor b2
• i.e. x3 = 0 AND x3 = 0 1 
• How about 111000  110011 001011 ?
– ??00??  ?100??  110001
• In general, we shall have:
– ??b??b??b?...?b  exact
– ??b??b??b?...?b?  exact
– ??b??b??b?...?b??  two cases …
27
C: Riding the Bus
28
C: Riding the Bus
• The order-3 curve contains 9 smaller copies of the order-2
curve (with 4 reversed left to right), connected by line
segments, as described for the order-2 curve. Higher order
curves are drawn in a similar manner.
• The connecting points to which processor components attach
are evenly spaced every len units along the bus. The first
connecting point is at (0,0) and the last is at (1,1). There are 9k
connecting points along the order-k curve, and the total bus
length is (9k–1)  len units.
• You must write a program to determine the total distance that
signals must travel between two processor components. Each
component’s coordinates are given as an x, y pair, 0  x  1
and 0  y  1, where x is the distance from the left side of the
chip, and y is the distance from the lower edge of the chip.
29
C: Riding the Bus
30
Riding the Bus: Solution
•
•
•
•
•
Recursive Structure, recursive formula
Locating the smallest cell containing the points
Find the nearest edge
Find the distances within cells
Find the distances between cells
31
D: Eurodiffusion
32
D: Eurodiffusion
• Member states decorate the coins with their own motif. No
matter which motif is on the coin, it can be used anywhere in
the 12 Member States.
• You must write a program to simulate the dissemination of
euro coins throughout Europe. Each city belongs to a country,
and a country is a rectangular part of the plane. Initially, each
city has one million (1000000) coins in its country’s motif.
Every day a representative portion of coins, based on the city’s
beginning day balance, is transported to each neighbor of the
city. A representative portion is defined as one coin for every
full 1000 coins of a motif.
• A city is complete when at least one coin of each motif is
present in that city. A country is complete when all of its cities
are complete. Your program must determine the time required
for each country to become complete.
33
D: Eurodiffusion
34
Eurodiffusion: Solution
• Simulation
• Graph Model: Vertex
Connectivity
• Iterative Flowing and
Testing
35
E: Covering Whole Holes
• Whether a given cover can
be used to completely cover
a specified hole. Covers and
holes are rectangular
polygons
• Moreover, both cover and
hole are aligned along the
same coordinate axes, and
are not supposed to be
rotated against each other —
just translated relative to
each other.
36
E: Covering Whole Holes
Input: the input consists of several descriptions of covers and holes. The first
line of each description contains two integers h and c (4  h  50 and 4  c
 50), the number of points of the polygon describing the hole and the
cover respectively. Each of the following h lines contains two integers x
and y, which are the vertices of the hole’s polygon in the order they would
be visited in a trip around the polygon. The next c lines give a
corresponding description of the cover. Both polygons are rectangular, and
the sides of the polygons are aligned with the coordinate axes. The
polygons have positive area and do not intersect themselves. The last
description is followed by a line containing two zeros.
Output: For each problem description, print its number in the sequence of
descriptions. If the hole can be completely covered by moving the cover
(without rotating it), print “Yes” otherwise print “No”. Recall that the cover
may extend beyond the boundaries of the hole as long as no part of the hole
is uncovered. Follow the output format in the example given below.
37
E: Covering Whole Holes
38
Covering Whole Holes: Solution
• Computational
Geometry
• PLAN: very few
numbers of edges
• Aligning the top edge
and the left edge
(double pairing)
• Testing overlaps
39
F: Combining Images
40
F: Combining Images
• The Quad Tree encoding of an image is a string of
binary digits. For easier printing, a Quad Tree
encoding can be converted to a Hex Quad Tree
encoding by the following steps:
• Prepend a 1 digit as a delimiter on the left of the
Quad Tree encoding.
• Prepend 0 digits on the left as necessary until the
number of digits is a multiple of four.
• Convert each sequence of four binary digits into a
hexadecimal digit, using the digits 0 to 9 and capital
A through F to represent binary patterns from 0000 to
1111.
41
F: Combining Images
• If an image is homogeneous (all its pixels are of the
same color), the Quad Tree encoding of the image is
1 followed by the color of the pixels. For example,
the Quad Tree encoding of an image that contains
pixels of color 1 only is 11, regardless of the size of
the image.
• If an image is heterogeneous (it contains pixels of
both colors), the Quad Tree encoding of the image is
0 followed by the Quad Tree encodings of its upperleft quadrant, its upper-right quadrant, its lower-left
quadrant, and its lower-right quadrant, in order.
42
Combining Images: Solution
•
•
•
•
•
Recurrence: The Quad Tree encoding.
PLAN: expand the formula to an n by n square array.
ANDing the two array
Converting the square array to a Quad Tree formula.
Encoding?
– 0abcd
– 1x
• Str2Tree ? Level? Tree2Mat(T, lev)? Mat2Tree, Tree2Str
43
G: A Linking Loader
• An object module contains (in order) zero or
more external symbol definitions, zero or more
external symbol references, zero or more bytes
of code and data (that may include references
to the values of external symbols), and an end
of module marker. In this problem, an object
module is represented as a sequence of text
lines, each beginning with a single uppercase
character that characterizes the remainder of
the line.
44
G: A Linking Loader
•
•
•
•
“D symbol offset” is an external symbol definition. Starting at the address 10016,
the line “D START 5C” indicates that the symbol START is defined as being
associated with the address 15C16. The number of D lines in a test case is at most
100.
“E symbol” is an external symbol reference, The line “E START” indicates that the
value of the symbol START. Each of the “E” lines for each module is numbered
sequentially, starting with 0, so they can be referenced in the “C” lines.
“C n byte1 byte2 … byten” specifies the first or next n bytes of code and data for
the current module. The value n is specified as a one or two digit hexadecimal
number, and will be no larger than 10 hexadecimal. Each byte is either a one or two
digit hexadecimal number, or a dollar sign. The values specified for the other bytes
(those not following a dollar sign) are loaded into sequential memory locations,
starting with the first (lowest) unused memory location. For example, the line “C 4
25 $ 0 37” would cause the values 2516 0116 5C16 and 3716 to be placed in the
next four unused memory locations, assuming the first “E” line for the current
module specified a symbol defined as having the address 15C16. If the 0-origin
index of the external symbol reference is an undefined symbol, the 16-bit value
inserted at the current point in the linked program is 000016.
A line of the form “Z” marks the end of an object module.
45
G: A Linking Loader
46
Linking Loader: Solution
• System Programming
– Tips: Read the problem carefully
– Know the meaning of every symbol
– How these values are packed / related to each other
• Symbol Table / Translation
• String Matching algorithm
47
H: A Spy in the Metro
48
H: A Spy in the Metro
• Maria knows that a powerful organization is after her.
She also knows that while waiting at a station, she is
at great risk of being caught. To hide in a running
train is much safer, so she decides to stay in running
trains as much as possible, even if this means
traveling backward and forward. Maria needs to
know a schedule with minimal waiting time at the
stations that gets her to the last station in time for her
appointment.
49
H: A Spy in the Metro
•
•
•
•
•
•
•
•
•
Input
The input file contains several test cases. Each test case consists of seven lines with
information as follows.
The integer N (2  N  50), which is the number of stations.
The integer T (0  T  200), which is the time of the appointment.
N-1 integers: t1, t2, ... tN-1 (1  ti  20), representing the travel times for the trains
between two consecutive stations: t1 represents the travel time between the first two
stations, t2 the time between the second and the third station, and so on.
The integer M1 (1  M1  50), representing the number of trains departing from the
first station.
M1 integers: d1, d2, ... dM1 (0  di  250 and di < di+1), representing the times at
which trains depart from the first station.
The integer M2 (1  M2  50), representing the number of trains departing from the
Nth station.
M2 integers: e1, e2, ... eM2 (0  ei  250 and ei < ei+1) representing the times at
which trains depart from the Nth station.
50
H: A Spy in the Metro
51
Spy in Metro: Solution
• Simulation / Dynamic Programming …
• (time, delay) calculations ; stop at the deadline
(34, 3)
0 (0, 0)
10
25
30
40
(15, 0)
(5, 0)
5
5
15
30
35
45
(36, 1)
10
15
25
40
45
55
(29, 0)
(23, 0)
8
11
26
36
56
5
20
30
50
6
52
I: The Solar System
53
I: The Solar System
The orbits obey three laws discovered by Johannes Kepler.
These laws, also called “The Laws of Planetary Motion,” are
the following.
1. The orbits of the planets are ellipses, with the sun at one focus
of the ellipse. (Recall that the two foci of an ellipse are such
that the sum of the distances to them is the same for all points
on the ellipse.)
2. The line joining a planet to the sun sweeps over equal areas
during equal time intervals as the planet travels around the
ellipse.
3. The ratio of the squares of the revolutionary periods of two
planets is equal to the ratio of the cubes of their semi major
axes.
54
I: The Solar System
• In this problem, you are to compute the location of a planet
using Kepler’s laws. You are given the description of one
planet in the Solar System (i.e., the length of its semi-major
axis, semi-minor axis, and its revolution time) and the
description of a second planet (its semi-major axis and semiminor axis).
• Assume that the second planet’s orbit is aligned with the
coordinate axes (as in the above figure), that it moves in
counter clockwise direction, and that the sun is located at the
focal point with non-negative x-coordinate.
• You are to compute the position of the second planet a
specified amount of time after it starts at the point with
maximal x-coordinate on its orbit (point B in the above figure).
55
I: The Solar System
• Input
• The input file contains several descriptions of pairs of planets.
Each line contains six integers a1, b1, t1, a2, b2, t. The first
five integers are positive, and describe two planets as follows:
• a1 = semi major axis of the first planet’s orbit
• b1 = semi minor axis of the first planet’s orbit
• t1 = period of revolution of the first planet (in days)
• a2 = semi major axis of the second planet’s orbit
• b2 = semi minor axis of the second planet’s orbit
• The non-negative integer t is the time (in days) at which you
have to determine the position of the second planet, assuming
that the planet starts in position (a2,0).
• The last description is followed by a line containing six zeros.
56
Solar System: Solution
• Compute the period. t2 = t1 *
sqrt(a23/a13)
• The planet starts in position (a2,0);
sun at (c, 0), c2+b2 = a2,why?
• Converting the time t = t mod t2
• A(x): The (upper half) ellipses
area, covered by the left focus;
– Numerical Integration
– Polar System: dA = r dθ/ 2
• Find the x such that A(x) = A(a2) *
t / t2
– Parametric (Binary) Searching
57
J: Toll
58
J: Toll
• Each town and village demanded an entry toll. The
toll for entering a village was simply one item. The
toll for entering a town was one piece per 20 items
carried. For example, to enter a town carrying 70
items, you had to pay 4 items as toll.
• Finding the cheapest route is a real challenge. The
best route depends upon the number of items carried.
For numbers up to 20, villages and towns charge the
same. For large numbers of items, it makes sense to
avoid towns and travel through more villages.
59
J: Toll
60
Toll: Solution
• Graph Searching + Dynamic Programming
• Backward Searching ; Cheapest Route?
76
72 71
70
66
42
11
12
10
45
39
44
13
12
11
43
42
61
ICPC2002 Kaohsiung: 8 Problems
62
E: Trees Isomorphism
63
Trees Isomorphism: Solution
• Find the root; Figure out the structure.
• Tree to Standard Form (String) of the Shape:
• Prefix Coding:
– T1  7 3 2 1 1 2 1 1 2 1 1 ;
– T2  7 2 2 1 1 4 2 1 2 1 1;
– T3  7 3 2 1 1 2 1 1 2 1 1 ;
64
F: Tree Size
65
F: Tree Size
i
k
j
• Consider nodes i, j, k
– Path length p(i, *) = (M[i][k]+M[i][j]-M[j][k])/2;
• Find the min p(i, *) by trying all possible j, k pair
• Shrink i
• Repeat the process until there is only one edge.
66
Final Remarks
•
•
Be brave, my friend, be brave.
Keep cool, How to solve it?
1.
2.
3.
4.
•
•
Understanding The Problem
Devising A Plan
Carrying Out The Plan
Looking Back
You can do it!
Downdload:
–
–
ICPC2003-World :http://icpc.baylor.edu/icpc/Finals/Problems.html
ICPC2002-Kaohsiung
http://icpc.baylor.edu/past/icpc2003/Regionals/Reports/icpc.baylor.e
du/icpc/regionals/Kaohsiun02/index.html
73
74