UVA Online Judge Solution - 101 - The blocks problem in C plus plus

UVA Online Judge Solution - 101 - The blocks problem in C plus plus 

Problem PDF is :

Many areas of Computer Science use simple, abstract domains for both analytical and empirical studies.

For example, an early AI study of planning and robotics (STRIPS) used a block world in which a robot

arm performed tasks involving the manipulation of blocks.

In this problem you will model a simple block world under certain rules and constraints. Rather

than determine how to achieve a specified state, you will “program” a robotic arm to respond to a

limited set of commands.

The problem is to parse a series of commands that instruct a robot arm in how to manipulate blocks

that lie on a flat table. Initially there are n blocks on the table (numbered from 0 to n􀀀1) with block

bi adjacent to block bi+1 for all 0 i < n 􀀀 1 as shown in the diagram below:

Initial Blocks World

The valid commands for the robot arm that manipulates blocks are:

• move a onto b

where a and b are block numbers, puts block a onto block b after returning any blocks that are

stacked on top of blocks a and b to their initial positions.

• move a over b

where a and b are block numbers, puts block a onto the top of the stack containing block b, after

returning any blocks that are stacked on top of block a to their initial positions.

• pile a onto b

where a and b are block numbers, moves the pile of blocks consisting of block a, and any blocks

that are stacked above block a, onto block b. All blocks on top of block b are moved to their

initial positions prior to the pile taking place. The blocks stacked above block a retain their order

when moved.

• pile a over b

where a and b are block numbers, puts the pile of blocks consisting of block a, and any blocks

that are stacked above block a, onto the top of the stack containing block b. The blocks stacked

above block a retain their original order when moved.

• quit

terminates manipulations in the block world.

Any command in which a = b or in which a and b are in the same stack of blocks is an illegal

command. All illegal commands should be ignored and should have no affect on the configuration of

blocks.

Input

The input begins with an integer n on a line by itself representing the number of blocks in the block

world. You may assume that 0 < n < 25.

The number of blocks is followed by a sequence of block commands, one command per line. Your

program should process all commands until the quit command is encountered.

You may assume that all commands will be of the form specified above. There will be no syntactically

incorrect commands.

Output

The output should consist of the final state of the blocks world. Each original block position numbered

i (0 i < n where n is the number of blocks) should appear followed immediately by a colon. If there

is at least a block on it, the colon must be followed by one space, followed by a list of blocks that appear

stacked in that position with each block number separated from other block numbers by a space. Don’t

put any trailing spaces on a line.

There should be one line of output for each block position (i.e., n lines of output where n is the

integer on the first line of input).

Sample Input

10

move 9 onto 1

move 8 over 1

move 7 over 1

move 6 over 1

pile 8 over 6

pile 8 over 5

move 2 over 1

move 4 over 9

quit

Sample Output

0: 0

1: 1 9 2 4

2:

3: 3

4:

5: 5 8 7 6

6:

7:

8:

9:

UVA Online Judge Solution - 101 - The blocks problem in C plus plus Code:

#include <iostream>
#include <cstdio>
using namespace std;
 
int a [30] [30];
int n;
int first_x;
int first_y;
int second_x;
int second_y;
 
void output ()
{
    for ( int i = 0; i < n; i++ ) {
        printf ("%d:", i);
        for ( int j = 0; a [i] [j] != -1; j++ )
            printf (" %d", a [i] [j]);
        printf ("\n");
    }
 
    //printf ("\n");
}
 
void constructor ()
{
    for ( int i = 0; i < n; i++ ) {
        a [i] [0] = i;
        a [i] [1] = -1;
    }
}
 
bool validity (int first, int second)
{
    if ( first == second )
    return false;
 
    for ( int i = 0; i < n; i++ ) {
        for ( int j = 0; a [i] [j] != -1; j++ ) {
            if ( first == a [i] [j] ) {
                for ( int k = 0; a [i] [k] != -1; k++ ) {
                    if ( second == a [i] [k] )
                        return false;
                }
            }
        }
    }
 
    return true;
}
 
 
void find (int first, int second)
{
    for ( int i = 0; i < n; i++ ) {
        for ( int j = 0; a [i] [j] != -1; j++ ) {
            if ( a [i] [j] == first ) {
                first_x = i;
                first_y = j;
            }
 
            else if ( a [i] [j] == second ) {
                second_x = i;
                second_y = j;
            }
        }
    }
}
 
void move_onto_method (int first, int second)
{
    if ( !validity (first, second) )
    return;
 
    find (first, second);
 
    //output ();
 
    for ( int i = first_y + 1; a [first_x] [i] != -1; i++ ) {
        a [a [first_x] [i]] [0] = a [first_x] [i];
        a [a [first_x] [i]] [1] = -1;
    }
 
    a [first_x] [first_y] = -1;
 
    for ( int i = second_y + 1; a [second_x] [i] != -1; i++ ) {
        a [a [second_x] [i]] [0] = a [second_x] [i];
        a [a [second_x] [i]] [1] = -1;
    }
 
    a [second_x] [second_y + 1] = first;
    a [second_x] [second_y + 2] = -1;
 
}
 
 
void move_over_method (int first, int second)
{
    if ( !validity (first, second) )
    return;
 
    find (first, second);
 
    for ( int i = first_y + 1; a [first_x] [i] != -1; i++ ) {
        a [a [first_x] [i]] [0] = a [first_x] [i];
        a [a [first_x] [i]] [1] = -1;
    }
 
    a [first_x] [first_y] = -1;
 
    second_y++;
    while ( a [second_x] [second_y] != -1 )
        second_y++;
 
    a [second_x] [second_y] = first;
    a [second_x] [++second_y] = -1;
}
 
 
void pile_onto_method (int first, int second)
{
    if ( !validity (first, second) )
    return;
 
    find (first, second);
 
    for ( int i = second_y + 1; a [second_x] [i] != -1; i++ ) {
        a [a [second_x] [i]] [0] = a [second_x] [i];
        a [a [second_x] [i]] [1] = -1;
    }
 
    for ( int i = first_y; a [first_x] [i] != -1; i++ )
        a [second_x] [++second_y] = a [first_x] [i];
 
    a [second_x] [++second_y] = -1;
    a [first_x] [first_y] = -1;
 
}
 
 
void pile_over_method (int first, int second)
{
    if ( !validity (first, second) )
    return;
 
    find (first, second);
 
    second_y++;
    while ( a [second_x] [second_y] != -1 )
        second_y++;
 
    for ( int i = first_y; a [first_x] [i] != -1; i++ )
        a [second_x] [second_y++] = a [first_x] [i];
 
    a [second_x] [second_y] = -1;
    a [first_x] [first_y] = -1;
}
 

int main ()
{
    
    while ( cin >> n ) {
 
        constructor ();
 
        
 
        string move_pile;
        int first;
        string onto_over;
        int second;
 
        while ( cin >> move_pile ) {
 
            if ( move_pile == "quit" )
                break;
            cin >> first;
            cin >> onto_over;
            cin >> second;
 
            if ( move_pile == "move" && onto_over == "onto" ) {
                move_onto_method (first, second);
            }
 
            else if ( move_pile == "move" && onto_over == "over" ) {
                move_over_method (first, second);
            }
 
            else if ( move_pile == "pile" && onto_over == "onto" ) {
                pile_onto_method (first, second);
            }
 
            else if ( move_pile == "pile" && onto_over == "over" ) {
                pile_over_method (first, second);
            }
 
            //output ();
        }
 
        output ();
    }
 
    return 0;
}

UVA Online Judge Solution 100 - 3n + 1 Problem solution in C plus plus

UVA Online Judge Solution 100 - 3n + 1 Problem solution in C plus plus 

UVA Online Judge Solution 100 - 3n + 1 Problem solution in C plus plus Problem is:

Problem PDF:

Problems in Computer Science are often classified as belonging to a certain class of problems (e.g.,
NP, Unsolvable, Recursive). In this problem you will be analyzing a property of an algorithm whose
classification is not known for all possible inputs.
Consider the following algorithm:
1. input n
2. print n
3. if n = 1 then STOP
4. if n is odd then n  􀀀 3n + 1
5. else n  􀀀 n/2
6. GOTO 2
Given the input 22, the following sequence of numbers will be printed
22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1
It is conjectured that the algorithm above will terminate (when a 1 is printed) for any integral input
value. Despite the simplicity of the algorithm, it is unknown whether this conjecture is true. It has
been verified, however, for all integers n such that 0 < n < 1; 000; 000 (and, in fact, for many more
numbers than this.)
Given an input n, it is possible to determine the number of numbers printed before and including
the 1 is printed. For a given n this is called the cycle-length of n. In the example above, the cycle
length of 22 is 16.
For any two numbers i and j you are to determine the maximum cycle length over all numbers
between and including both i and j.
Input
The input will consist of a series of pairs of integers i and j, one pair of integers per line. All integers
will be less than 10,000 and greater than 0.
You should process all pairs of integers and for each pair determine the maximum cycle length over
all integers between and including i and j.
You can assume that no operation overflows a 32-bit integer.
Output
For each pair of input integers i and j you should output i, j, and the maximum cycle length for
integers between and including i and j. These three numbers should be separated by at least one space
with all three numbers on one line and with one line of output for each line of input. The integers i
and j must appear in the output in the same order in which they appeared in the input and should be
followed by the maximum cycle length (on the same line).
Sample Input
1 10
100 200
201 210
900 1000
Sample Output
1 10 20
100 200 125
201 210 89
900 1000 174

UVA Online Judge Solution 100 - 3n + 1 Problem solution in C plus plus 

#include <stdio.h>
#include <string.h>
#define max(x, y) ((x) > (y) ? (x) : (y))
#define swap(x, y) {int tmp; tmp = x, x = y, y = tmp;}
#define N 1000000
#define node 1048576
int A[N+1], tree[node<<1];
void setTree(int l, int r, int k) {
    if(l == r)
        tree[k] = A[l];
    if(l < r) {
        int m = (l+r) >> 1;
        setTree(l, m, k<<1);
        setTree(m+1, r, (k<<1)+1);
        tree[k] = max(tree[k<<1], tree[(k<<1)+1]);
    }
}
int getTree(int l, int r, int k, int i, int j) {
    if(l == i && r == j)
        return tree[k];
    if(l < r) {
        int m = (l+r) >> 1;
        if(m >= j)
            return getTree(l, m, k<<1, i, j);
        else if(m < i)
            return getTree(m+1, r, (k<<1)+1, i, j);
        else
            return max(getTree(l, m, k<<1, i, m), getTree(m+1, r, (k<<1)+1, m+1, j));
    }
}
void build() {
    long long n;
    int i, len;
    memset(A, 0, sizeof(A));
    memset(tree, 0, sizeof(tree));
    A[1] = 1;
    for(i = 2; i <= N; i++) {
        n = i, len = 1;
        while(n != 1) {
            if(n&1) {
                n = n*3 + 1;
            } else {
                n >>= 1;
            }
            if(n < N && A[n] != 0) {
                len += A[n];
                break;
            }
            len ++;
        }
        A[i] = len;
    }
    setTree(1, N, 1);
}
int main() {
    build();
    int i, j;
    while(scanf("%d %d", &i, &j) == 2) {
        printf("%d %d ", i, j);
        if(i > j)
            swap(i, j);
        printf("%d\n", getTree(1, N, 1, i, j));
    }
    return 0;
}