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Closed 11 years ago.

m 试图在一线至最后一线之间找到最短的道路。问题始终是我的法典,我感觉到这一点,因为它把第一线和第一线之间的距离计算出来,后者将达到零,但不会达到100%。我的法典为什么总是回到零?

特设矩阵是[10][10],所有节点都连接起来,例如网点是矩阵。

private static int dijkstras(Graph g) {
    // Dijkstra s Algorithm
    int[] best = new int[g.network.length];
    boolean[] visited = new boolean[g.network.length];
    int max = 10000; // Infinity equivalent.
    for (int i = 0; i < g.network.length; i++)
    {
        best[i] = max;
        visited[i] = false;
    }

    best[0] = 0;

    for(int i = 0; i < g.network.length; i++)
    {
        int min = max;
        int currentNode = 0;
        for (int j = 0; j < g.network.length; j++)
        {
            if (!visited[j] && best[j] < min)
            {
                currentNode = j;
                min = best[j];
            }
        }
        visited[currentNode] = true;
        for (int j = 0; j < g.network.length; j++)
        {
            if (g.network[currentNode][j] < max && best[currentNode] + g.network[currentNode][j] < best[j])
            {
                best[j] = best[currentNode] + g.network[currentNode][j];
            }
        }
    }
            return best[g.network.length - 2];
}

Answer 1

我认为,我也许已经通过修改以下法典来解决这个问题(现在,在起点而不是“0”中硬编码的一点):

    private static int dijkstras(Graph g, int start) // Added a start point.
    {
    // Dijkstra s Algorithm
    int[] best = new int[g.network.length];
    boolean[] visited = new boolean[g.network.length];
    int max = 10000; // Infinity equivalent.
    for (int i = 0; i < g.network.length; i++)
    {
        best[i] = max;
        visited[i] = false;
    }

    best[start] = start; // Changed the 0 to variable start.

    for(int i = 0; i < g.network.length; i++)
    {
        int min = max;
        int currentNode = 0;
        for (int j = 0; j < g.network.length; j++)
        {
            if (!visited[j] && best[j] < min)
            {
                currentNode = j;
                min = best[j];
            }
        }
        visited[currentNode] = true;
        for (int j = 0; j < g.network.length; j++)
        {
            if (g.network[currentNode][j] < max && best[currentNode] +   g.network[currentNode][j] < best[j])
            {
                best[j] = best[currentNode] + g.network[currentNode][j];
            }
        }
    }
            return best[g.network.length - 2];
}

我通过 lo和沃伊拉来实施这一法典,现在还不是零。

Answer 2

我读过你的法典,我几乎肯定这句话是正确的分法算法,因此,我认为,在您的第一点和最后一点之间,你也许会走零路。

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