给定图邻接矩阵(例如[] [] []),指向图.
需求找到所有图形周期的计数(如果存在)并打印它们.

我试图用Java编写这个算法,有时它可以正常工作.如果图形具有复杂的周期,则算法返回疯狂周期.请查看我的代码并帮助解决此问题

public static final int k = 6;

public static int g[][] = { { 0, 1, 0, 0, 0, 0 },
                            { 1, 0, 1, 0, 0, 0 },
                            { 0, 0, 0, 1, 0, 0 },
                            { 0, 0, 0, 0, 1, 0 },
                            { 0, 0, 1, 0, 0, 0 },
                            { 0, 0, 0, 0, 0, 0 } };

public static Vector stack = new Vector();

public static void printStack() {
    System.out.print("stack is: { ");
    for (int i = 0; i < stack.size(); i++) {
        System.out.print(stack.get(i) + " ");
    }
    System.out.println("};");

}

public static boolean checkCycle() {
    boolean res = false;

    for (int i = 0; i < stack.size() - 1; i++) {
        if (stack.get(i).equals(stack.lastElement())) {
            res = true;
            break;
        }

    }
    return res;
}

public static boolean go_to_line(int line) {
    boolean res = false;
    for (int i = 0; i < k; i++) {
        if (g[line][i] == 1) {
            stack.add(i);
            if (checkCycle() == true) {
                System.out.println("Cycle found!");
                res = true;
            } else {
                res = go_to_line(i);
            }
        }
    }

    return res;
}

public static int cycles_count() {
    int res = 0;

    for (int i = 0; i < k; i++) {
        if (g[i][i] == 1) {
            System.out.println("Knot detected at item {" + i + "}!");
            res++;
        }

        for (int j = i + 1; j < k; j++) {
            if (g[j][i] == 1) {
                stack.add(j);
                stack.add(i);

                if (go_to_line(i) == true) {
                    res++;

                    System.out.print("Final ");
                    printStack();
                    stack.removeAllElements();
                }
            }
        }
    }

    return res;
}

解决方法:

在一般情况下,该问题具有指数复杂性.问题在于,如果每个顶点连接到每个顶点,则所有图形循环的计数大于2 ^ n(节点的任何子集形成几个循环).

因此,在一般情况下没有好的算法.
要找到某个周期,您可以使用广度优先搜索.要查找所有周期,您应该使用强力算法.