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我正在为一个编程竞赛练习,在这个竞赛中,我可以选择使用 Python 还是 C++ 来解决每个问题,因此我对任何一种语言的解决方案都持开放态度——无论哪种语言最适合这个问题。

我坚持的过去问题的 URL 是http://progconz.elena.aut.ac.nz/attachments/article/74/10%20points%20Problem%20Set%202012.pdf,问题 F(“地图”) .

基本上,它涉及在一个大的 ASCII 艺术中匹配出现的一小块 ASCII 艺术。在 C++ 中,我可以为每幅 ASCII 艺术作品制作一个矢量。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。我不想要为我编写的所有代码,只是对问题所需逻辑的想法。

谢谢你的帮助。

这是我到目前为止所得到的:

#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. 它有效,但我不明白这一切。我已经评论了我可以但需要帮助来理解函数return中的语句count_pattern

#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() )
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3 回答 3

1

不要用线条来思考。将整个页面读入一个字符串,并像对待任何其他字符一样处理行尾字符。

(您可能认为这是一个神秘的提示,但您确实只是要求“一个想法”如何做到这一点。)

编辑:由于您知道图片的整体尺寸,您可以从您尝试匹配的模式的第一行向前计数字符以匹配第二行,依此类推以匹配后续行。

于 2013-06-30T02:13:35.743 回答
1
#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 )
)

生成器表达式可以使用显式的 for 循环块重写:

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。

构建子矩阵以与模式进行比较的列表推导可以优化为更早返回:

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

或者使用显式的 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
于 2013-06-30T06:26:11.187 回答
0
#include <iostream>
#include <vector>

using namespace std;

int main(){

    int numOfRounds;
    cin >> numOfRounds;



    for(int round = 0; round < numOfRounds; round++){

        int out = 0;

        int linesToMatch;
        cin >> linesToMatch;

        int sizeToMatch;
        cin >> sizeToMatch;

        vector <string> v;
        string t;

        for (int i = 0; i < linesToMatch; i++){
            cin >> t;
            v.push_back(t);
        }

        string s = "";

        int rows;
        cin >> rows;

        int columns;
        cin >> columns;

        for (int j = 0; j < rows; j++){        //map->string
            cin >> t;
            s += t;
        }

        // main part of implementation
        // it's mainly basic algebra and index tracking
        for (int m = 0; m <= rows - linesToMatch; m++){
            for (int n = 0; n <= columns - sizeToMatch; n++){
                int x;
                for (x = 0; x < linesToMatch; x++){
                    int index = (m + x) * columns + n;
                    string newTemp(s.begin() + index, s.begin() + index + sizeToMatch);
                    if (newTemp != v.at(x)) break;
                }
                if (x == linesToMatch) out++;
            }
        }

        cout << out << endl;

    }

}
于 2013-06-30T02:59:25.800 回答