我一直在尝试用Java实现Floyd-Warshall算法而不使用“三循环嵌套”方式，但我似乎无法弄清楚我在代码中出错的地方。 / p>
这是显示我的顶点如何连接的地图。白色数字是顶点，黑色数字是连接顶点之间的距离。
顶点地图：http://i.imgur.com/htcaA4y.png
运行迭代后，我得到以下最终距离和序列矩阵。说“出错了”的是最终序列矩阵的第8列（右边的那个）。为了从任何其他顶点到达顶点8，路径必须首先从顶点8变为9，然后变为10（根据矩阵不是这种情况 - 它从顶点8直接变为10）。
输出矩阵：http://i.imgur.com/o6fQweH.png
这是代码。什么似乎是问题？
import java.util.ArrayList;
public class Main_Simple {
public static void main(String[] args) {
// 1. Setup the distance matrix
// -inf for vertices that are not connected
// -### for edge weights of vertices that are connected
// -0 across the diagonal
int inf = 1000000; // Temporary 'infinity' variable
// Initial distance matrix must be n x n
int[][] initialDistanceMatrix = {
{0, 1, inf, 1, inf, inf, inf, inf, inf, inf, inf, inf, inf, inf, inf},
{1, 0, 1, inf, 1, inf, inf, inf, inf, inf, inf, inf, inf, inf, inf},
{inf, 1, 0, inf, inf, 1, inf, inf, inf, inf, inf, inf, inf, inf, inf},
{1, inf, inf, 0, 1, inf, inf, inf, inf, inf, inf, inf, inf, inf, inf},
{inf, 1, inf, 1, 0, 1, inf, inf, inf, inf, inf, inf, inf, inf, inf},
{inf, inf, 1, inf, 1, 0, 2, inf, inf, 1, inf, inf, inf, inf, inf},
{inf, inf, inf, inf, inf, 2, 0, inf, inf, inf, inf, inf, inf, inf, inf},
{inf, inf, inf, inf, inf, inf, inf, 0, 1, inf, inf, inf, inf, inf, inf},
{inf, inf, inf, inf, inf, inf, inf, 1, 0, 1, inf, inf, inf, inf, inf},
{inf, inf, inf, inf, inf, 1, inf, inf, 1, 0, 2, 1, inf, inf, inf},
{inf, inf, inf, inf, inf, inf, inf, inf, inf, 2, 0, inf, inf, inf, 2},
{inf, inf, inf, inf, inf, inf, inf, inf, inf, 1, inf, 0, 1, inf, inf},
{inf, inf, inf, inf, inf, inf, inf, inf, inf, inf, inf, 1, 0, 1, inf},
{inf, inf, inf, inf, inf, inf, inf, inf, inf, inf, inf, inf, 1, 0, 1},
{inf, inf, inf, inf, inf, inf, inf, inf, inf, inf, 2, inf, inf, 1, 0}
};
// 2. Setup the sequence matrix
// -All of column-1 are ones
// -All of column-2 are twos
// -etc
// -0 across the diagonal
// Initial sequence matrix must be the same size as the initial distance matrix
int[][] initialSequenceMatrix = new int[initialDistanceMatrix.length][initialDistanceMatrix.length];
for (int row = 0; row < initialSequenceMatrix.length; row++) {
for (int column = 0; column < initialSequenceMatrix.length; column++) {
if (row == column) {
initialSequenceMatrix[row][column] = 0;
} else {
initialSequenceMatrix[row][column] = column + 1; // +1 to account 0-based array
}
}
}
// 3. Iterate through the matrices (n-1) times
// -On the kth iteration, copy the kth column and kth row down to the next distance and sequence matrix
// -On the kth iteration, check matrix (k-1) and take the minimum of the following two:
// -d(ij)
// -d(ik)+d(kj)
// where i = row number, j = column number, and k = iteration number
// -After the distance matrix has been calculated, compare the current distance matrix to the previous.
// If the numbers are the same, keep the sequence matrix the same. Otherwise, change the sequence
// matrix to the current iteration's number.
ArrayList<int[][]> distanceMatrices = new ArrayList<int[][]>();
distanceMatrices.add(initialDistanceMatrix);
ArrayList<int[][]> sequenceMatrices = new ArrayList<int[][]>();
sequenceMatrices.add(initialSequenceMatrix);
// Print the matrices to make sure they are made correctly
printMatrix(initialDistanceMatrix, "Initial distance matrix");
printMatrix(initialSequenceMatrix, "Initial sequence matrix");
// Matrix Iteration Loops
for (int iteration = 1; iteration < initialDistanceMatrix.length; iteration++) {
// Initialize new distance matrix
int[][] currentDistanceMatrix = new int[initialDistanceMatrix.length][initialDistanceMatrix.length];
for (int row = 0; row < currentDistanceMatrix.length; row++) {
for (int column = 0; column < currentDistanceMatrix.length; column++) {
currentDistanceMatrix[row][column] = 0;
} // ends 'column' loop
} // ends 'row' loop
// Distance Matrix iteration
for (int row = 0; row < currentDistanceMatrix.length; row++) {
for (int column = 0; column < currentDistanceMatrix.length; column++) {
if (row == column) { // If you are on the diagonal, insert '0'
currentDistanceMatrix[row][column] = 0;
} else if (row == (iteration - 1) || column == (iteration - 1)) { // Brings down the row and column of the iteration (-1 to account 0-based array)
currentDistanceMatrix[row][column] = distanceMatrices.get(iteration - 1)[row][column];
} else { // If you are on any other square...
int Dij = distanceMatrices.get(iteration - 1)[row][column];
int Dik_Dkj = distanceMatrices.get(iteration - 1)[row][iteration - 1] + distanceMatrices.get(iteration - 1)[iteration - 1][column];
if (Dij > Dik_Dkj) currentDistanceMatrix[row][column] = Dik_Dkj;
else currentDistanceMatrix[row][column] = Dij;
}
} // ends 'column' loop
} // ends 'row' loop
// Add the distance matrix to the matrix array
distanceMatrices.add(currentDistanceMatrix);
// Initialize new sequence matrix
int[][] currentSequenceMatrix = new int[initialDistanceMatrix.length][initialDistanceMatrix.length];
// Sequence Matrix iteration
for (int row = 0; row < currentSequenceMatrix.length; row++) {
for (int column = 0; column < currentSequenceMatrix.length; column++) {
if (row == column) { // If you are along the diagonal...
currentSequenceMatrix[row][column] = 0;
} else if (row == (iteration - 1) || column == (iteration - 1)) { // If you are on the same row or column as the iteration...
currentSequenceMatrix[row][column] = sequenceMatrices.get(iteration - 1)[row][column];
} else { // If you are on any other square...
// You need to check the current distance matrix to see if it matches the previous.
// If it does match, keep the same number.
// If it changed, changed the number in that cell to the current iteration
// Compare the most recent distance matrix to the one before it
if (distanceMatrices.get(distanceMatrices.size() - 1)[row][column] == distanceMatrices.get(distanceMatrices.size() - 2)[row][column]) {
currentSequenceMatrix[row][column] = sequenceMatrices.get(sequenceMatrices.size() - 1)[row][column];
} else {
currentSequenceMatrix[row][column] = iteration;
}
}
} // ends 'column' loop
} // ends 'row' loop
// Add the sequence matrix to the matrix array
sequenceMatrices.add(currentSequenceMatrix);
} // ends matrix iteration loops
System.out.println("-------------------------------------------------------");
printMatrix(distanceMatrices.get(distanceMatrices.size() - 1), "Final Distance Matrix");
printMatrix(sequenceMatrices.get(sequenceMatrices.size() - 1), "Final Sequence Matrix");
} // ends main method
public static void printMatrix(int[][] matrix, String message) {
System.out.println("\n" + message);
for (int row = 0; row < matrix.length; row++) {
for (int column = 0; column < matrix.length; column++) {
System.out.print(matrix[row][column] + "\t");
} // ends 'column' loop
System.out.println();
} // ends 'row' loop
System.out.println();
}
} // ends class Main_Simple
答案 0 :(得分：0)
你没有在第一个循环中正确地遍历所有顶点。
let writeNumbersInFile (file : string) (numbers : int seq) =
use outFile = new StreamWriter (file)
Seq.iter (string >> outFile.WriteLine) numbers
outFile.Flush ()
[1;2;3;4]
|> writeNumbersInFile "4.txt"
应该是：
for (int iteration = 1; iteration < initialDistanceMatrix.length; iteration++)
或者，更好的是，不要将for (int iteration = 1; iteration < initialDistanceMatrix.length + 1; iteration++)
用于所有数组索引，而是使用iteration - 1，然后您可以使用：
iteration这使得所有索引都保持为零而不是混合它们。只有这个代码更改（以及一个更小的测试用例），我得到了正确的答案。