Classificar uma lista com trocas e pops

21

Considere uma lista aleatória dos números inteiros 1 a N. Você deseja classificá-la usando apenas as seguintes ações:

  1. Troque o primeiro e o último elementos da lista. (S)
  2. Retire o primeiro elemento e anexe-o ao final da lista. (P)

Isso sempre é possível porque qualquer lista pode ser classificada com trocas suficientes de elementos adjacentes. Usando S e P, você pode trocar qualquer elemento adjacente chamando P até que os dois elementos em questão sejam o primeiro e o último item da lista, depois chamando S para trocá-los e depois chamando P novamente até que estejam nos índices originais (agora trocados )

No entanto, em termos do número de operações S e P, é improvável que esse método seja ideal para a maioria das listas.

Desafio

Escreva um programa ou função que permita uma permutação dos números de 1 a N (N> 1). Pode ser dada como uma lista ou sequência ou o que for conveniente. Ele deve gerar uma sequência de S e P que classifique a permutação quando aplicada da esquerda para a direita. Essa sequência não precisa ser idealmente curta, mas, quanto menor, melhor (consulte Pontuação).

Exemplo

Se a entrada foi [2, 1, 3]a saída pode ser SPporque

  • S aplicados para [2, 1, 3]o torna [3, 1, 2],
  • e P aplicado a [3, 1, 2]faz [1, 2, 3], que é classificado.

Verificador

Esse snippet de pilha pode ser usado para verificar se uma sequência realmente classifica a lista. A lista deve ter colchetes e ser separada por vírgula. A sequência SP é apenas uma sequência de S'e P'.

<style>*{font-family:monospace}</style><script>function swap(list) {var tmp = list[0];list[0] = list[list.length - 1];list[list.length - 1] = tmp;}function pop(list) {list.push(list.shift())}function check() {var result = 'Sorted sucessfully.';var details = '';try {var list = eval(document.getElementById('list').value);var seq = document.getElementById('seq').value;details += 'Sequence: ' + seq + '<br>Steps:<br>' + list + ' <-- original';for(var i = 0; i < seq.length; i++) {if (seq.charAt(i) == 'S')swap(list);else if (seq.charAt(i) == 'P')pop(list);details += ' (' + i + ')<br>' + list;}details += ' <-- final (' + i + ')';for(var i = 1; i < list.length; i++) {if (list[i-1] > list[i]) {result = 'Sorting failed!';break;}}} catch(e) {result = 'Error: ' + e;}document.getElementById('result').innerHTML = result;document.getElementById('details').innerHTML = details;}</script><p>List<br><input id='list'type='text'size='60'value='[2, 1, 3]'></p><p>SP Sequence<br><textarea id='seq'rows='1'cols='60'>SP</textarea><br></p><button type='button'onclick='check()'>Check</button><p id='result'></p><p id='details'></p>

Pontuação

Seu algoritmo deve produzir seqüências SP corretas, mas possivelmente subótimas, para N = 2 a 256. Para qualquer permutação nesse intervalo, ele deve ser executado em menos de 5 minutos em um computador moderno decente .

Sua pontuação é o número total de operações S e P que seu algoritmo precisa para classificar todas as 30 listas listadas abaixo. A pontuação mais baixa vence.

Você não pode codificar seu programa de acordo com esses dados. Se parecer necessário, posso adicionar mais dados de teste para determinar o vencedor.

1. Length 3
[2, 1, 3]
2. Length 7
[2, 7, 5, 3, 4, 6, 1]
3. Length 41
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4. Length 52
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5. Length 65
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6. Length 69
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7. Length 75
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8. Length 80
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9. Length 103
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10. Length 108
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11. Length 109
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12. Length 151
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13. Length 173
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14. Length 177
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15. Length 192
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16. Length 201
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17. Length 201
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18. Length 205
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19. Length 211
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20. Length 226
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21. Length 227
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22. Length 230
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23. Length 238
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24. Length 241
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25. Length 241
[105, 160, 132, 32, 10, 117, 76, 2, 190, 108, 178, 51, 71, 237, 232, 47, 14, 124, 100, 31, 169, 196, 8, 184, 21, 151, 223, 86, 42, 127, 55, 58, 229, 4, 219, 46, 238, 179, 24, 194, 203, 122, 66, 15, 81, 146, 172, 106, 129, 135, 216, 120, 92, 231, 144, 195, 181, 162, 69, 45, 137, 136, 9, 30, 5, 188, 91, 49, 147, 233, 198, 17, 241, 163, 36, 18, 183, 59, 16, 29, 116, 182, 41, 48, 23, 39, 154, 210, 68, 167, 95, 213, 79, 225, 37, 157, 109, 143, 78, 142, 173, 155, 200, 110, 20, 73, 141, 168, 156, 126, 150, 201, 114, 1, 230, 211, 217, 131, 140, 204, 209, 149, 103, 199, 165, 175, 35, 107, 74, 63, 193, 239, 218, 234, 197, 224, 174, 121, 60, 88, 22, 171, 133, 207, 152, 34, 43, 228, 125, 115, 101, 7, 12, 220, 82, 153, 134, 52, 130, 70, 72, 44, 177, 89, 65, 98, 94, 53, 208, 227, 161, 3, 123, 236, 221, 87, 102, 50, 90, 180, 185, 186, 67, 77, 26, 212, 27, 222, 6, 145, 11, 13, 84, 25, 164, 64, 85, 139, 214, 189, 61, 96, 191, 128, 93, 138, 148, 19, 119, 202, 75, 176, 159, 56, 104, 206, 40, 187, 215, 62, 166, 240, 112, 226, 170, 83, 97, 57, 99, 38, 28, 113, 205, 158, 235, 111, 54, 192, 118, 80, 33]
26. Length 244
[89, 13, 154, 161, 235, 225, 92, 188, 215, 194, 54, 58, 128, 21, 165, 85, 144, 205, 142, 77, 109, 24, 83, 69, 72, 7, 224, 240, 191, 204, 183, 203, 68, 70, 63, 95, 206, 170, 153, 180, 45, 178, 35, 27, 190, 132, 222, 41, 40, 156, 97, 20, 217, 177, 167, 65, 23, 136, 216, 234, 38, 201, 236, 164, 82, 241, 10, 141, 148, 229, 5, 125, 113, 159, 193, 187, 130, 179, 52, 108, 86, 196, 174, 123, 56, 116, 227, 30, 239, 98, 186, 67, 135, 118, 163, 43, 32, 50, 231, 226, 232, 172, 200, 99, 6, 143, 39, 93, 107, 34, 129, 157, 100, 127, 15, 84, 81, 73, 121, 220, 44, 8, 57, 105, 91, 171, 162, 211, 244, 104, 112, 238, 212, 133, 71, 192, 145, 160, 87, 181, 60, 184, 119, 4, 55, 96, 53, 12, 213, 124, 94, 22, 42, 3, 48, 131, 117, 140, 138, 228, 219, 155, 59, 29, 202, 169, 114, 101, 47, 233, 176, 139, 207, 33, 79, 16, 51, 66, 75, 90, 198, 168, 166, 31, 151, 49, 208, 150, 111, 14, 25, 197, 17, 76, 80, 78, 9, 173, 26, 137, 19, 120, 199, 106, 152, 88, 147, 36, 158, 28, 243, 221, 110, 74, 175, 237, 61, 2, 62, 214, 189, 134, 195, 218, 18, 102, 46, 103, 11, 122, 146, 185, 182, 209, 1, 149, 115, 64, 37, 126, 230, 223, 242, 210]
27. Length 246
[28, 186, 214, 18, 220, 73, 20, 88, 234, 187, 27, 102, 22, 129, 10, 228, 196, 56, 47, 2, 16, 67, 124, 137, 177, 179, 223, 147, 188, 23, 103, 109, 149, 60, 229, 99, 222, 90, 49, 80, 158, 93, 171, 71, 175, 143, 19, 212, 40, 226, 219, 120, 146, 66, 167, 232, 94, 174, 237, 9, 173, 70, 122, 241, 58, 82, 191, 211, 180, 104, 53, 36, 83, 37, 131, 25, 162, 32, 210, 144, 145, 69, 135, 63, 154, 165, 46, 57, 50, 74, 128, 140, 112, 118, 125, 231, 221, 233, 95, 200, 153, 6, 166, 98, 208, 91, 89, 141, 4, 26, 134, 86, 8, 30, 157, 156, 224, 201, 243, 7, 161, 1, 84, 115, 44, 230, 78, 79, 5, 17, 194, 148, 152, 121, 193, 107, 240, 181, 29, 43, 65, 164, 45, 110, 64, 195, 216, 127, 59, 197, 178, 151, 139, 85, 75, 11, 204, 184, 77, 54, 51, 205, 108, 142, 130, 138, 238, 87, 38, 101, 96, 31, 215, 244, 242, 172, 213, 136, 97, 35, 39, 76, 42, 245, 123, 227, 24, 168, 33, 159, 217, 239, 55, 176, 68, 207, 106, 132, 15, 116, 61, 198, 62, 182, 225, 202, 105, 150, 183, 81, 170, 13, 41, 199, 3, 185, 235, 14, 155, 21, 126, 119, 169, 72, 12, 236, 48, 209, 190, 113, 218, 100, 52, 111, 189, 92, 192, 114, 117, 160, 133, 203, 163, 34, 206, 246]
28. Length 250
[119, 57, 59, 181, 212, 120, 236, 121, 99, 247, 138, 187, 175, 108, 107, 197, 123, 101, 141, 77, 201, 3, 52, 60, 56, 240, 157, 39, 42, 199, 23, 18, 136, 74, 137, 143, 229, 170, 20, 160, 206, 219, 191, 185, 46, 223, 150, 190, 116, 96, 198, 221, 220, 159, 238, 48, 176, 113, 168, 33, 44, 142, 67, 244, 13, 218, 122, 246, 214, 234, 237, 27, 165, 24, 153, 90, 84, 154, 235, 196, 80, 111, 102, 231, 228, 135, 16, 148, 26, 45, 10, 71, 156, 224, 183, 232, 72, 94, 132, 54, 58, 248, 144, 213, 139, 129, 245, 115, 25, 164, 87, 19, 193, 81, 32, 78, 91, 167, 171, 40, 92, 226, 109, 69, 86, 38, 61, 5, 14, 118, 145, 103, 53, 93, 172, 178, 225, 68, 163, 210, 179, 192, 208, 169, 17, 12, 50, 233, 6, 55, 243, 158, 88, 188, 242, 36, 173, 126, 155, 184, 216, 149, 47, 76, 200, 1, 112, 28, 161, 174, 41, 73, 222, 66, 37, 63, 64, 124, 89, 205, 9, 186, 202, 70, 203, 127, 105, 250, 182, 79, 43, 133, 8, 241, 114, 128, 51, 83, 98, 49, 209, 7, 95, 151, 162, 189, 180, 75, 195, 62, 207, 21, 104, 30, 117, 110, 140, 29, 227, 249, 82, 152, 11, 34, 85, 106, 217, 211, 2, 35, 215, 4, 230, 134, 31, 194, 97, 22, 125, 130, 100, 239, 131, 166, 65, 15, 146, 177, 204, 147]
29. Length 253
[46, 245, 174, 180, 85, 29, 141, 70, 252, 119, 214, 225, 86, 91, 41, 67, 219, 118, 127, 243, 2, 71, 157, 114, 55, 92, 5, 200, 199, 139, 191, 235, 153, 102, 206, 171, 117, 58, 223, 249, 11, 211, 202, 175, 156, 133, 57, 163, 47, 65, 213, 247, 189, 111, 177, 49, 124, 154, 164, 145, 80, 15, 232, 142, 39, 69, 201, 185, 229, 215, 170, 42, 3, 248, 169, 136, 149, 218, 237, 90, 135, 7, 242, 246, 23, 186, 31, 4, 167, 207, 173, 132, 195, 196, 78, 212, 22, 35, 194, 54, 184, 183, 222, 230, 88, 43, 144, 151, 34, 97, 6, 109, 37, 147, 72, 158, 134, 33, 51, 238, 165, 162, 9, 63, 166, 113, 137, 198, 50, 121, 106, 224, 19, 104, 95, 129, 193, 79, 192, 122, 30, 12, 234, 168, 76, 103, 73, 66, 188, 178, 172, 176, 250, 182, 110, 231, 155, 26, 197, 10, 152, 98, 131, 25, 36, 81, 138, 150, 227, 24, 112, 120, 204, 75, 8, 179, 94, 17, 140, 32, 77, 61, 233, 38, 205, 93, 99, 27, 208, 56, 216, 48, 241, 20, 130, 240, 115, 83, 60, 108, 28, 13, 89, 128, 160, 82, 40, 126, 16, 253, 21, 251, 228, 59, 159, 64, 203, 236, 52, 18, 181, 105, 107, 239, 220, 44, 74, 125, 209, 84, 226, 217, 210, 190, 101, 100, 68, 116, 161, 148, 244, 221, 96, 45, 123, 187, 62, 87, 53, 1, 146, 143, 14]
30. Length 256
[159, 248, 75, 43, 111, 38, 4, 17, 155, 87, 81, 208, 53, 230, 65, 11, 108, 228, 146, 212, 137, 225, 144, 189, 86, 105, 84, 128, 97, 50, 223, 15, 83, 169, 217, 47, 88, 236, 114, 181, 115, 177, 102, 250, 246, 104, 80, 45, 240, 14, 196, 52, 247, 41, 198, 32, 182, 206, 226, 101, 70, 94, 113, 49, 254, 59, 42, 154, 77, 253, 112, 215, 99, 25, 134, 92, 95, 150, 64, 178, 118, 79, 130, 63, 129, 131, 57, 218, 85, 204, 3, 163, 158, 186, 10, 199, 24, 125, 251, 51, 74, 160, 207, 120, 233, 30, 19, 9, 56, 167, 58, 117, 164, 54, 220, 229, 234, 203, 211, 103, 205, 69, 39, 5, 31, 191, 106, 180, 116, 245, 21, 222, 91, 235, 8, 2, 214, 29, 238, 176, 135, 61, 227, 202, 122, 252, 231, 89, 187, 37, 149, 96, 190, 172, 73, 18, 71, 1, 179, 46, 156, 201, 165, 256, 151, 27, 242, 188, 126, 124, 244, 100, 107, 143, 170, 72, 255, 40, 67, 192, 185, 219, 20, 157, 98, 237, 136, 168, 141, 224, 232, 44, 139, 132, 22, 60, 109, 90, 175, 171, 62, 23, 161, 12, 76, 210, 68, 184, 93, 243, 48, 209, 174, 200, 121, 123, 36, 173, 140, 133, 145, 6, 110, 152, 142, 241, 55, 138, 147, 193, 213, 127, 7, 34, 66, 153, 26, 13, 162, 183, 239, 78, 249, 221, 28, 194, 16, 35, 33, 82, 195, 148, 216, 119, 166, 197]
Passatempos de Calvin
fonte
Este é um desafio interessante. Mas a questão é: precisamos calcular manualmente a pontuação ou você fará isso em uma tabela de classificação?
Optimizer
1
@Optimizer Prefiro que você calcule sua pontuação, só para não precisar me preocupar com cada nova resposta ou edição. Vou executar os programas para confirmar as melhores pontuações quando selecionar o vencedor.
Passatempos de Calvin
Eis algumas casos de teste adicional: [2,1], [2,1,4,3], [1,2,6,4,5,3]. Apenas no caso de pessoas estão tentando coisas mais complicadas :)
SP3000

Respostas:

9

Python 3 (com PyPy) - pontuação: 607771

Edit: Eu acho que o bug foi corrigido, mas ainda não estou convencido de que isso funcione até que eu prove que ele funciona

def d(a, b, N):
    a, b = a%N, b%N

    if a >= b:
        return a-b
    else:
        return a+(N-b)

def gt(a, b, N):
    if a == N and b == 1:
        return False
    if a == 1 and b == N:
        return True
    return a > b

def solve(numbers):
    N = len(numbers)
    best_move = None
    target = sorted(numbers)

    if N == 2:
      return "" if numbers == [1,2] else "P"

    for b in range(N):
        nums = numbers[:]
        moves = []
        head_ptr = 0 # Which element should be at the start of the list if we actually moved stuff
        reference = target[b:]+target[:b]
        ref_d = {t:i for i,t in enumerate(reference)}

        while True:
            for pops in range(N):
                ptr = (head_ptr + pops) % N

                # Magic
                d1 = d(ref_d[nums[ptr-1]],ptr-1,N)+d(ptr,ref_d[nums[ptr]],N)
                d2 = d(ref_d[nums[ptr]],ptr,N)+d(ptr-1,ref_d[nums[ptr-1]],N)

                if d1 > d2 or (d1 == d2 and gt(nums[ptr-1], nums[ptr], N)):
                    moves.append("P"*pops + "S")
                    head_ptr = (head_ptr + pops) % N
                    nums[ptr-1],nums[ptr] = nums[ptr],nums[ptr-1]
                    break

            else:
                if nums[nums.index(1):]+nums[:nums.index(1)] != target:
                    print("This algorithm is seriously broken.\n")
                moves.append("P"*((N+nums.index(1)-head_ptr)%N))
                break

        moves = "".join(moves)

        if not best_move or len(moves) < len(best_move):
            best_move = moves

    return best_move

Uso

solve([3, 4, 2, 1, 5])

Explicação

O programa apenas realiza trocas de classificação por bolhas, usando a observação de que:
S: n 1 2 3 ... n-1 0 PS: 0 2 3 ... n-1 n 1 PPS: 1 3 ... n-1 n 0 2 PPPS: 2 ... n-1 n 0 1 3 ...

Em outras palavras, os ipushs seguidos por uma troca significam que o elemento ié trocado por elemento i-1, quebra de índices. No entanto, precisamos acompanhar o fato de que a lista é alterada ao fazê-lo.

Para fazer melhor do que apenas o tipo de bolha usual, tentamos aplicar uma heurística para garantir que não façamos muitos swaps inúteis, reforçando que qualquer swap aproxime os dois elementos de seus respectivos objetivos (não provei formalmente está correto, mas parece funcionar até agora). Depois, tiramos a melhor solução de muitas configurações (o CPython leva um tempo, mas o PyPy faz esse trabalho com relativa rapidez).

Pontuação

Saída em Pastebin.


Ver classificação anterior de bolha ingênua. (pontuação: 1126232)

Este definitivamente funciona.

def cyclic_sorted(nums):
    for i in range(len(nums)):
        if nums[i:] + nums[:i] == sorted(nums):
            return i
    return False

def solve(nums):
    N = len(nums)

    def gt(a, b):
        if a == 1 and b == N:
            return True
        if a == N and b == 1:
            return False
        return a > b

    moves = []
    head_ptr = 0 # Which element should be at the start of the list if we actually moved stuff

    while nums != sorted(nums):
        pops = 0
        ptr_offset = 0

        while pops < N:
            ptr = (head_ptr + pops) % N

            if (not moves or pops > 0) and gt(nums[ptr-1], nums[ptr]):
                moves.append("P"*pops + "S")
                head_ptr = (head_ptr + pops) % N

                nums[ptr-1],nums[ptr] = nums[ptr],nums[ptr-1]
                break

            pops += 1

        else:
            moves.append("P"*cyclic_sorted(nums[head_ptr:]+nums[:head_ptr]))
            break

    return "".join(moves)
Sp3000
fonte
3
É interessante que isso forme um padrão, mesmo que os dados sejam aleatórios. Eu transformei P em espaços para sua lista de 256 (role perto da parte inferior).
Passatempos de Calvin
É interessante que sua implementação tenha uma pontuação melhor que a minha, embora, lendo sua descrição, basicamente façamos o mesmo. Possivelmente há alguma diferença que esteja faltando (ainda não li o seu código completamente). ^^
ReyCharles
8

Javascript ES6 - 607.771

function popsort( iData ) {
    var sSP, iLen,
        sSP_Opt = popsortimpl( iData.slice( 0 ), 0 ),
        iOptLen = sSP_Opt.length;

    for( var n = iData.length, i = 2; i <= n; i++ )
        if( (iLen = (sSP = popsortimpl( iData.slice( 0 ), i - 1 )).length) < iOptLen ) {
            iOptLen = iLen;
            sSP_Opt = sSP;
        }

    return sSP_Opt;
}

function popsortimpl( iData, iPivIdx ) {
    function issorted()
        { return iData.every( (_,i_) => _ == i_+1 ); }

    function swap() {
        var i_ = iData[0];

        iData[0] = iData[n-1];
        iData[n-1] = i_;

        sSP.push( 'S' );
    }

    function pop() {
        iData.push( iData.shift() );
        if( --iPivIdx < 0 )
            iPivIdx += n;

        sSP.push( 'P' );
    }

    function dist2home( iIdx_, i_ ) {
        var iHome_ = (iPivIdx + i_) %n,
            iDist1_ = iIdx_ - iHome_,
            iDist2_ = iHome_ - iIdx_;

        if( iDist1_ < 0 )
            iDist1_ += n;
        if( iDist2_ < 0 )
            iDist2_ += n;

        return Math.min( iDist1_, iDist2_*Math.max( 1, n/50 ) );
    }

    var n = iData.length,
        sSP = [];

    while( !issorted() ) {
        var iDistP = Math.max( dist2home( 0, iData[0] ), dist2home( n - 1, iData[n - 1] ) ),
            iDistS = Math.max( dist2home( n - 1, iData[0] ), dist2home( 0, iData[n - 1] ) );

        if( iDistS < iDistP )
            swap();
        else pop();
    }
    return sSP.join('');
}

Implementa uma classificação de bolha "mais próxima" que borbulha elementos para cima ou para baixo, dependendo da rota mais curta (bolha para cima ou para baixo) até uma posição de referência.

Não acredito que a rotina seja ótima, mas deve estar razoavelmente próxima. Como bônus, ele executa o conjunto completo de vetores de teste em milissegundos.

Atualizada para iterar sobre o ponto de articulação para melhorar a competitividade. Eu também introduzi heurísticas para compensar o fato de que os elementos têm uma mobilidade menor subindo "a lista" do que movendo-se "para baixo" (um fenômeno análogo à diferença na mobilidade de elétrons e buracos em um semicondutor). Existem grandes diferenças entre essa implementação e o Sp3000, portanto, o fato de ambos terem exatamente 607.771 pontos é assustador. Dia das Bruxas assustador.

A rotina agora leva cerca de 20 segundos para ser executada.

Saída de amostra

popsort( [2, 7, 5, 3, 4, 6, 1] )

rendimentos

PSPPSPSPPPS
COTO
fonte
4

OCaml, pontuação: 1136707 1136599

Edit: Eu errei! Eu tinha lido Strocaria elementos adjacentes e perdi que o primeiro e último elementos trocados. Essa implementação se baseia nisso, como é óbvio a partir da evalimplementação. Editar²: Eu consertei usando a observação de que você pode escrever o que eu pensava ser SP como P S. Também mudei para a versão com melhor pontuação com algumas otimizações para torná-la menos lenta.

O tempo de execução é de 1,33 18 1,01 segundos no meu sistema usando o binário compilado intérprete .

type sp = S | P

let string_of_sp = function
  | S -> "S"
  | P -> "P"

let rec eval (e : sp list) (xs : int list) : int list =
  match e, xs with
  | [], xs -> xs
  | P :: S :: e, x1 :: x2 :: xs -> eval e (x1 :: xs @ [x2])
  | P :: e, x :: xs -> eval e (xs @ [x])
  | _ -> assert false

let generate_sp_sequence xs : sp list =
  let sorted = List.sort compare xs in
  let rec generate_sp_sequence ((xs, ys) : int list * int list) acc : sp list =
    if sorted = ys @ List.rev xs
    then acc
    else 
      match ys with
      | v1 :: v2 :: ys' ->
         if v1 > v2
         then generate_sp_sequence (v2 :: xs, v1 :: ys') (S :: P :: acc)
         else generate_sp_sequence (v1 :: xs, v2 :: ys') (P :: acc)
      | [x] ->
         let ys = List.rev (x :: xs) in
         generate_sp_sequence ([], ys) (P :: acc)
      | [] ->
         assert false
  in List.rev @@ generate_sp_sequence ([], xs) []


(** Uncomment to run tests **)
(*
let test =
  QCheck.(mk_test
            ~n:1000
            ~name:"generate_sp_sequence sorting test"
            ~pp:PP.(list int)
            Arbitrary.(list ~start:2 ~stop:256 small_int)
            (fun xs ->
             Prop.assume (List.length xs >= 2);
             eval (generate_sp_sequence ([], xs))
                  xs
             = List.sort compare xs))

let _ = QCheck.run test
 *)

let score_test = ... (* not included for brevity *)

let score =
  List.map (fun xs -> generate_sp_sequence ([], xs)) score_test
  |> List.map List.length
  |> List.fold_left (+) 0

let _ =
  print_string "The score is: ";
  print_int score;
  print_newline ()

generate_sp_sequence simplesmente executa bolhas e produz as etapas em termos de S e P. Isso pode gerar etapas desnecessárias.

ReyCharles
fonte
4

C, pontuação: 607771

Acontece que minha solução é muito semelhante à solução Python do Sp3000 (que explica a mesma pontuação). Eu nunca usei Python, então não posso dizer com certeza, mas acho que o conceito é exatamente o mesmo:

Para cada número r em 0..n-1, defina uma meta de uma matriz classificada rotacionada r vezes a partir da posição inicial (ou seja, contendo P pops com r = P mod n) e encontre uma série de movimentos resultando nessa configuração. Grave a menor seqüência de movimentos para qualquer r.
Realize swaps apenas quando o swap aproximar-se de qualquer elemento que esteja mais distante da posição do objetivo (olhando em qualquer direção) ou quando o swap aproxima os dois elementos se ambos estiverem igualmente distantes.

Código

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

/* Returns true if the array is sorted starting at position 'offset' */
int is_sorted(int *array, int n, int offset)
{
    int i;
    for (i = 0; i < n && array[(offset+i)%n] == i+1; i++);
    return i == n;
}

/* Find moves that would sort the array, assuming the first element will
   be at position (offset + n/2) % n */
int get_moves(int *array, int n, int offset, int stop_after, char *moves)
{
    int tmp, posA = n-1, posB = 0, counter = 0;
    /* Repeat until sorted or reaching the maximum number of moves (previous
       best, or the worst case upper bound if no previous solution) */
    for (; counter < stop_after && !is_sorted(array, n, posB); posA = (posA+1)%n, posB = (posB+1)%n, moves[counter++] = 'P')
    {
        /* Perform a swap if it would move whichever element is further away
           closer to its target */
        if ((array[posA] - posA + offset + n) % n >
            (array[posB] - posB + offset + n) % n)
        {
            tmp = array[posA];
            array[posA] = array[posB];
            array[posB] = tmp;
            moves[counter++] = 'S';
            if (is_sorted(array, n, posB))
            {
                break;
            }
        }
    }
    return counter;
}

int *parse_input(int _argc, char **_argv)
{
    int i;
    int *array = NULL;
    if (_argc > 1)
    {
        array = malloc((_argc-1) * sizeof(int));
        _argv[1]++; /* remove initial square bracket */
        for (i = 1; i < _argc; i++)
        {
            array[i-1] = atoi(_argv[i]);
        }
    }
    return array;
}

int main(int _argc, char **_argv)
{
    int *orig_array = parse_input(_argc, _argv);
    if (orig_array != NULL)
    {
        int n = _argc - 1;
        /* Safe worst case upper bound representing n^2 "SP" moves */
        int worst_move_count = 2*n*n;

        int i, move_count, offset;
        int best_offset, best_move_count = worst_move_count;

        char *best_moves = malloc(worst_move_count + 1);
        char *moves = malloc(worst_move_count + 1);
        int *array = malloc(n * sizeof(int));

        memset(best_moves, 0, worst_move_count + 1);
        /* Try all offsets from 0 to n-1; i.e. find solutions for:
           1, 2, 3, ..., n-2, n-1, n
           2, 3, 4, ..., n-1, n,   1
           3, 4, 5, ..., n,   1,   2
           ...
           n, 1, 2, ..., n-3, n-2, n-1
        */
        for (offset = 0; offset < n; offset++)
        {
            memset(moves, 0, worst_move_count + 1);
            memcpy(array, orig_array, n*sizeof(int));
            move_count = get_moves(array, n, offset, best_move_count, moves);
            if (move_count < best_move_count)
            {
                strcpy(best_moves, moves);
                best_move_count = move_count;
                best_offset = offset;
            }
        }
        printf("%s\n", best_moves);

        free(best_moves);
        free(moves);
        free(array);
        free(orig_array);
    }
    return 0;
}

Uso

$ gcc -Ofast -o popswap popswap.c
$ ./popswap [2, 7, 5, 3, 4, 6, 1]
PSPPSPSPPPS
$ ./popswap [2, 7, 5, 3, 4, 6, 1]|tr -d '\n'|wc -c
11

Pontuação

Roteiro:

#!/bin/sh
grep ^\\[ input.txt|while read line; do
    len=`echo $line|wc -w`;
    score=`./popswap $line|tr -d '\n'|wc -c`;
    echo "    $len $score";
done

Resultados:

$ ./scoring.sh |cut -d' ' -f 6|paste -sd+|bc
607771
$ time ./scoring.sh > /dev/null

real    0m1.926s
user    0m1.879s
sys     0m0.122s
$ ./scoring.sh
3 2
7 11
41 893
52 1546
65 2543
69 2686
75 3391
80 3728
103 6483
108 0
109 7103
151 14371
173 18663
177 19728
192 22824
201 24828
201 24752
205 25573
211 27509
226 32221
227 31928
230 33630
238 34603
241 36676
241 36505
244 37759
246 38106
250 38264
253 40040
256 41405
Mike Patterson
fonte
1

C ++ 11, pontuação: 1122052

Implementação muito ingênua que não leva em consideração os buracos na sequência de entrada, mas leva em consideração o valor mínimo na sequência.

A lista de argumentos é apenas a sequência de números, sem vírgulas e colchetes (por exemplo: "./a.out 2 7 5")

#include <iostream>
#include <cstring>
#include <string>
using namespace std;

int* input  = 0;
int s_count = 0;
int p_count = 0;
int len;
int last;

void do_swap()
{
  int a = input[0];
  input[0] = input[last];
  input[last] = a;
  s_count++;
  cerr << "S";
}

void do_pop()
{
  int a = input[0];
  memmove(&input[0], &input[1], sizeof(int) * last);
  input[last] = a;
  p_count++;
  cerr << "P";
}

bool is_sorted()
{
  int i = 0;
  while (input[i] < input[i+1]) { if (++i == last) return true; }
  return false;
}

int main(int argc, char** args)
{
  int min_val = 256;

  len = argc - 1;
  last = len - 1;
  if (len <= 1)
    return 1;

  input = new int[len];
  for (int i = 2; i <= argc; i++)
  {
    int j = stoi(args[i-1]);
    input[i-2] = j;
    min_val = min(min_val, j);
  }

  while (!is_sorted())
  {
    if ((input[0]) == min_val)
      do_pop();
    else if (input[0] < input[last])
      do_swap();
    else if (input[0] > input[last])
      do_pop();
  }

  cerr << endl;
  cout << len << " " << (s_count + p_count) << endl;
  delete [] input;
  return 0;
}

Pontuação

3 2
7 19
41 1642
52 3236
65 4787
69 4966
75 6644
80 6090
103 12142
108 0
109 13774
151 24976
173 34061
177 37835
192 41994
201 48316
201 47276
205 49439
211 51551
226 58859
227 56824
230 58994
238 64227
241 65774
241 66287
244 71345
246 73426
250 71766
253 68629
256 77171
BlakBat
fonte