Perancangan dan Analisis Algoritme Lanjut

Agus Budi Raharjo

5109100164

Jurusan Teknik Informatika

Fakultas Teknologi Informasi Institut Teknologi Sepuluh Nopember

Latihan

• SPOJ Problem Set (classical) 140 – The Loner

• SPOJ Problem Set (tutorial) 67 – Cutting out • SPOJ Problem Set (tutorial) 404 – Scanner

SPOJ 140 The Loner

SPOJ Problem Set (classical)

140. The Loner

Problem code: LONER

The loner is a one-dimensional board game for a single player. The board is composed of squares arranged in a

single line, some of which initially have pawns on them. The player makes a move by jumping with a pawn over

a pawn on an adjacent field, to an empty square two fields to the right or left of its initial position. The pawn that

was jumped over is removed directly after the move, as illustrated below.

The game is considered won if exactly one pawn remains on the gaming board, and is lost if the player cannot

make a move.

Given the initial state of the gaming board, your task is to determine whether it is possible for the player to win

the game.

SPOJ 140 The Loner

Input

The input begins with the integer t, the number of test cases. Then t test cases follow. Each test cases begins with

the positive integer n <= 32000, denoting the size of the gaming board. The second and last line of the test case

description contains a sequence of n characters 0 or 1, without any white spaces. The i-th square of the board is

occupied by a pawn at the start of the game iff the i-th character of this sequence is 1.

Output

For each test case output the word yes if it is possible for the player to win the game for the presented starting

configuration, or the word no in the opposite case.

Example

Sample input:

2

7

0110011

6

111001

Sample output:

yes

no

SPOJ 140 The Loner

SPOJ 140 The Loner

SPOJ 140 The Loner

Langkah penyelesaian : 1. Hilangkan angka 0 di depan dan di belakang

permainan utama (0*L0*)

2. Untuk special case, definisikan

3. Cari persamaan antara array dengan teorema 1

SPOJ 140 The Loner

Dari teorema 1 didapat kemungkinan menang susunan permainan (L; dengan 0*L0* ):

• 1 • 011 • 110 • 11(01)*00(10)*11 • 11(01)*00(11)+(10)*11 • 11(01)*(11) +00(10)*11 • 11(01)*(11)*1011(10)*11 • 11(01)*1101(11)*(10)*11 • 11(01)*(11)*01 • 10 (11)*(10)*11

Ket:

tanda WX*Y artinya WY, WXY, WXXY, WX.....Y

tanda WX+Y artinya WXY, WXXY, WXXXY, WX.....Y

SPOJ 140 The Loner

untuk mengubahnya menjadi bentuk kondisional (menggunakan if), kita jabarkan dalam bentuk tree :

Ket : tanda garis bawah dan huruf “f” menandakan akhir array

SPOJ 140 The Loner

• Cont’d

Ket : tanda garis bawah dan huruf “f” menandakan akhir array

SPOJ 67 Cutting Out

SPOJ Problem Set (classical)

67. Cutting out

Problem code: CUTOUT

One has to cut out a number of rectangles from a paper square. The sides of each rectangle are to be parallel to the

sides of the square. Some rectangles can be already cut out. What is the largest area of a rectangle which can be cut

out from the remaining paper?

Illustration

Three rectangles have been cut out from the square 10x10 in the figure shown below. The area of the largest

rectangle that can be cut out from the remaining paper is 16. One of such rectangles is shown with a dashed line.

Task

Write a program that for each data set from a sequence of several data sets:

reads descriptions of a square and rectangles from the input,

computes the area of the largest rectangle which can be cut out from the remaining paper,

writes the result to output.

SPOJ 67 Cutting Out

Langkah penyelesaian : 1. Membuat Array persegi dengan sisi sesuai test

case dan membuat status “true”

2. Untuk setiap perpotongan, objek hasil perpotongan berganti status menjadi “false”

3. Untuk mencari luas persegi terbesar cek tiap cell, jika terdapat cell bernilai true, maka akan dicek luas perseginya dan dibandingkan dengan luas sebelumnya

SPOJ 67 Cutting Out

Alur pemeriksaan : 0 : posisi sekarang 1 : cek panjang (+1) 2 : cek lebar(sesuai dengan panjang) 3 : jika sudah membentuk persegi, index akan menuju ke cell 4 4 : cek panjang baru (+1) 5 : mencari lebar 6 : jika diketahui terdapat perpotongan, maka luas yang digunakan

adalah luas sebelumnya didapat dari index sekarang -1 dikuadratkan