Algorithme hongrois: comment couvrir 0 éléments avec des lignes minimales?

cochez la 3ème étape de l'algorithme dans l'article Wikipedia (section interprétation matricielle ) , ils expliquent une façon de calculer la quantité minimale de lignes pour couvrir tous les 0

mise à Jour: La suite est une autre façon d'obtenir le nombre minimum de lignes qui couvrent le 0's :

import java.util.ArrayList;
import java.util.List;

public class MinLines { 
    enum LineType { NONE, HORIZONTAL, VERTICAL }

    private static class Line {
        int lineIndex;
        LineType rowType;
        Line(int lineIndex, LineType rowType) { 
            this.lineIndex = lineIndex;
            this.rowType = rowType;
        }      
        LineType getLineType() {
            return rowType;
        }

        int getLineIndex() { 
            return lineIndex; 
        }
        boolean isHorizontal() {
            return rowType == LineType.HORIZONTAL;
        }
    }

    private static boolean isZero(int[] array) {
        for (int e : array) {
            if (e != 0) {
                return false;
            }
        }
        return true;
    }

    public static List<Line> getMinLines(int[][] matrix) {
        if (matrix.length != matrix[0].length) {
            throw new IllegalArgumentException("Matrix should be square!");
        }

        final int SIZE = matrix.length;
        int[] zerosPerRow = new int[SIZE];
        int[] zerosPerCol = new int[SIZE];

        // Count the number of 0's per row and the number of 0's per column        
        for (int i = 0; i < SIZE; i++) { 
            for (int j = 0; j < SIZE; j++) { 
                if (matrix[i][j] == 0) { 
                    zerosPerRow[i]++;
                    zerosPerCol[j]++;
                }
            }
        }

        // There should be at must SIZE lines, 
        // initialize the list with an initial capacity of SIZE
        List<Line> lines = new ArrayList<Line>(SIZE);

        LineType lastInsertedLineType = LineType.NONE;

        // While there are 0's to count in either rows or colums...
        while (!isZero(zerosPerRow) && !isZero(zerosPerCol)) { 
            // Search the largest count of 0's in both arrays
            int max = -1;
            Line lineWithMostZeros = null;
            for (int i = 0; i < SIZE; i++) {
                // If exists another count of 0's equal to "max" but in this one has
                // the same direction as the last added line, then replace it with this
                // 
                // The heuristic "fixes" the problem reported by @JustinWyss-Gallifent and @hkrish
                if (zerosPerRow[i] > max || (zerosPerRow[i] == max && lastInsertedLineType == LineType.HORIZONTAL)) {
                    lineWithMostZeros = new Line(i, LineType.HORIZONTAL);
                    max = zerosPerRow[i];
                }
            }

            for (int i = 0; i < SIZE; i++) {
                // Same as above
                if (zerosPerCol[i] > max || (zerosPerCol[i] == max && lastInsertedLineType == LineType.VERTICAL)) {
                    lineWithMostZeros = new Line(i, LineType.VERTICAL);
                    max = zerosPerCol[i];
                }
            }

            // Delete the 0 count from the line 
            if (lineWithMostZeros.isHorizontal()) {
                zerosPerRow[lineWithMostZeros.getLineIndex()] = 0; 
            } else {
                zerosPerCol[lineWithMostZeros.getLineIndex()] = 0;
            }

            // Once you've found the line (either horizontal or vertical) with the greater 0's count
            // iterate over it's elements and substract the 0's from the other lines 
            // Example:
            //                          0's x col:
            //           [ 0  1  2  3 ]  ->  1
            //           [ 0  2  0  1 ]  ->  2
            //           [ 0  4  3  5 ]  ->  1
            //           [ 0  0  0  7 ]  ->  3
            //             |  |  |  | 
            //             v  v  v  v
            // 0's x row: {4} 1  2  0 

            //           [ X  1  2  3 ]  ->  0
            //           [ X  2  0  1 ]  ->  1
            //           [ X  4  3  5 ]  ->  0
            //           [ X  0  0  7 ]  ->  2
            //             |  |  |  | 
            //             v  v  v  v
            //            {0} 1  2  0 

            int index = lineWithMostZeros.getLineIndex(); 
            if (lineWithMostZeros.isHorizontal()) {
                for (int j = 0; j < SIZE; j++) {
                    if (matrix[index][j] == 0) {
                        zerosPerCol[j]--;
                    }
                }
            } else {
                for (int j = 0; j < SIZE; j++) {
                    if (matrix[j][index] == 0) {
                        zerosPerRow[j]--;
                    }
                }                    
            }

            // Add the line to the list of lines
            lines.add(lineWithMostZeros); 
            lastInsertedLineType = lineWithMostZeros.getLineType();
        }
        return lines;
    }

    public static void main(String... args) { 
        int[][] example1 = 
        { 
           {0, 1, 0, 0, 5},
           {1, 0, 3, 4, 5},
           {7, 0, 0, 4, 5},
           {9, 0, 3, 4, 5},
           {3, 0, 3, 4, 5} 
        };

        int[][] example2 = 
        {
           {0, 0, 1, 0},
           {0, 1, 1, 0},
           {1, 1, 0, 0},
           {1, 0, 0, 0},
        };

        int[][] example3 = 
        {
           {0, 0, 0, 0, 0, 0},
           {0, 0, 0, 1, 0, 0},
           {0, 0, 1, 1, 0, 0},
           {0, 1, 1, 0, 0, 0},
           {0, 1, 0, 0, 0, 0},
           {0, 0, 0, 0, 0, 0}
        };

        List<int[][]> examples = new ArrayList<int[][]>();
        examples.add(example1);
        examples.add(example2);
        examples.add(example3);

        for (int[][] example : examples) {
            List<Line> minLines = getMinLines(example);
            System.out.printf("Min num of lines for example matrix is: %d\n", minLines.size());
            printResult(example, minLines);
            System.out.println();
        }
    }

    private static void printResult(int[][] matrix, List<Line> lines) {
        if (matrix.length != matrix[0].length) {
            throw new IllegalArgumentException("Matrix should be square!");
        }

        final int SIZE = matrix.length;
        System.out.println("Before:");
        for (int i = 0; i < SIZE; i++) {
            for (int j = 0; j < SIZE; j++) {
                System.out.printf("%d ", matrix[i][j]);
            }
            System.out.println();
        }

        for (Line line : lines) {
            for (int i = 0; i < SIZE; i++) {
                int index = line.getLineIndex();
                if (line.isHorizontal()) {
                    matrix[index][i] = matrix[index][i] < 0 ? -3 : -1;
                } else {
                    matrix[i][index] = matrix[i][index] < 0 ? -3 : -2;
                }
            }
        }   

        System.out.println("\nAfter:");
        for (int i = 0; i < SIZE; i++) {
            for (int j = 0; j < SIZE; j++) {
                System.out.printf("%s ", matrix[i][j] == -1 ? "-" : (matrix[i][j] == -2 ? "|" : (matrix[i][j] == -3 ? "+" : Integer.toString(matrix[i][j]))));
            }
            System.out.println();
        }   
    }
}   

la partie importante est la méthode getMinLines , il renvoie une List avec les lignes qui couvrent les entrées de la matrice 0's . Pour l'exemple matrices imprime:

Min num of lines for example matrix is: 3
Before:
0 1 0 0 5 
1 0 3 4 5 
7 0 0 4 5 
9 0 3 4 5 
3 0 3 4 5 

After:
- + - - - 
1 | 3 4 5 
- + - - - 
9 | 3 4 5 
3 | 3 4 5 

Min num of lines for example matrix is: 4
Before:
0 0 1 0 
0 1 1 0 
1 1 0 0 
1 0 0 0 

After:
| | | | 
| | | | 
| | | | 
| | | | 

Min num of lines for example matrix is: 6
Before:
0 0 0 0 0 0 
0 0 0 1 0 0 
0 0 1 1 0 0 
0 1 1 0 0 0 
0 1 0 0 0 0 
0 0 0 0 0 0 

After:
- - - - - - 
- - - - - - 
- - - - - - 
- - - - - - 
- - - - - - 
- - - - - -    

j'espère que cela vous donnera un coup de pouce, le reste de l'algorithme hongrois ne devrait pas être difficile à mettre en œuvre""