c++ - 如何在ASCII艺术作品中匹配ASCII艺术作品片段？

我正在参加编程竞赛，在该竞赛中，我将针对每个问题选择使用Python还是C++，因此我愿意使用任何一种语言(无论哪种语言最适合该问题)进行解决方案。

我遇到的过去问题的URL是http://progconz.elena.aut.ac.nz/attachments/article/74/10%20points%20Problem%20Set%202012.pdf，问题F(“ map ”)。

基本上，它涉及将一小块ASCII艺术作品匹配到一个大的艺术作品中。在C++中，我可以为每个ASCII艺术作品制作一个 vector 。问题是当较小的部分是多行时如何匹配它。

我不知道该怎么做。我不想为我编写所有代码，而只是想知道问题所需要的逻辑。

谢谢你的帮助。

到目前为止，这是我得到的:

#include <cstdlib>
#include <iostream>
#include <string>
#include <algorithm>
#include <vector>

using namespace std;

int main( int argc, char** argv )
{
    int nScenarios, areaWidth, areaHeight, patternWidth, patternHeight;

    cin >> nScenarios;

    for( int a = 0; a < nScenarios; a++ )
    {
        //get the pattern info and make a vector
        cin >> patternHeight >> patternWidth;
        vector< vector< bool > > patternIsBuilding( patternHeight, vector<bool>( patternWidth, false ) );

        //populate data
        for( int i = 0; i < patternHeight; i++ )
        {
            string temp;
            cin >> temp;
            for( int j = 0; j < patternWidth; j++ )
            {
                patternIsBuilding.at( i ).at( j ) = ( temp[ j ] == 'X' );
            }
        }

        //get the area info and make a vector
        cin >> areaHeight >> areaWidth;
        vector< vector< bool > > areaIsBuilding( areaHeight, vector<bool>( areaWidth, false ) );

        //populate data
        for( int i = 0; i < areaHeight; i++ )
        {
            string temp;
            cin >> temp;
            for( int j = 0; j < areaWidth; j++ )
            {
                areaIsBuilding.at( i ).at( j ) = ( temp[ j ] == 'X' );
            }
        }


        //now the vectors contain a `true` for a building and a `false` for snow
        //need to find the matches for patternIsBuilding inside areaIsBuilding
        //how?

    }


    return 0;
}

编辑:从下面的评论中，我从J.F. Sebastian获得了Python中的解决方案。它有效，但我不了解所有情况。我已经评论了我可以做的，但是需要帮助您了解return函数中的count_pattern语句。

#function to read a matrix from stdin
def read_matrix():

    #get the width and height for this matrix
    nrows, ncols = map( int, raw_input().split() )

    #get the matrix from input
    matrix = [ raw_input() for _ in xrange( nrows ) ]

    #make sure that it all matches up
    assert all(len(row) == ncols for row in matrix)

    #return the matrix
    return matrix

#perform the check, given the pattern and area map
def count_pattern( pattern, area ):

    #get the number of rows, and the number of columns in the first row (cause it's the same for all of them)
    nrows = len( pattern )
    ncols = len( pattern[0] )

    #how does this work?
    return sum(
        pattern == [ row[ j:j + ncols ] for row in area[ i:i + nrows ] ]
        for i in xrange( len( area ) - nrows + 1 )
        for j in xrange( len( area[i] ) - ncols + 1 )
    )

#get a number of scenarios, and that many times, operate on the two inputted matrices, pattern and area
for _ in xrange( int( raw_input() ) ):
    print count_pattern( read_matrix(), read_matrix() )

最佳答案

#how does this work?
return sum(
    pattern == [ row[ j:j + ncols ] for row in area[ i:i + nrows ] ]
    for i in xrange( len( area ) - nrows + 1 )
    for j in xrange( len( area[i] ) - ncols + 1 )
)

可以使用显式的for循环块重写生成器表达式:

count = 0
for i in xrange( len( area ) - nrows + 1 ):
    for j in xrange( len( area[i] ) - ncols + 1 ):
        count += (pattern == [ row[ j:j + ncols ]
                              for row in area[ i:i + nrows ] ])
return count

比较(pattern == ..)返回True/False，在Python中等于1/0。

可以优化构建子矩阵以与模式进行比较的列表理解，以使其更早返回:

count += all(pattern_row == row[j:j + ncols]
             for pattern_row, row in zip(pattern, area[i:i + nrows]))

或使用显式的for循环块:

for pattern_row, row in zip(pattern, area[i:i + nrows]):
    if pattern_row != row[j:j + ncols]:
       break # no match (the count stays the same)
else: # matched (no break)
    count += 1 # all rows are equal

关于c++ - 如何在ASCII艺术作品中匹配ASCII艺术作品片段？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/17386458/