8 皇帝问题(皇帝还是比较牛逼) - 算法

版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/sunyanxiong123/article/details/75331797

一、问题描述
有一个 8 * 8 的棋盘,和 8 个皇后,皇后的攻击规则:任意一个皇后可以攻击同一行、同一列、正反对角线上的皇后。问题:如何摆放8个皇后,可以让她们互相不攻击。
提示:一行或一列只能放一个皇后。

二、解题思路
采用深度优先遍历方式,从第0行开始,逐行查看每一列中那些位置可以摆放皇后,也就是每一行需要扫描8列,而且还需要判断当前位置和摆放过皇后的位置是否符合规则。

三、DFS(深度优先搜索)
套路:

// 参数n:问题规模
void dfs(int n){
// dfs 出口
 if(){
 ...
 return;
 }

// dfs body
处理当前内容

// dfs 递归
处理当前内容与子问题的关系

}

二叉树遍历

package com.daxiong.day6;

/**
 *          1
 *      2       3
 *    4   5   6   7
 * */
public class TowTree {

    private static int[] tree = {0,1,2,3,4,5,6,7};  // 第0为站位,不参与实际问题

    public static void main(String[] args) {
        dfs(1);
    }

    // 二叉树前序遍历
    public static void dfs(int root){

        if(root*2 >= 8){
            System.out.println(tree[root]);
            return;
        }
        System.out.println(tree[root]);
        dfs(root*2);   // 左子树
        dfs(root*2+1); // 右子树
    }

}

四、代码实现

package com.daxiong.day6;

/**
 * 8 皇后问题
 * */
public class EightQueen {

    // 8 个皇后
    private static int SIZE = 8;
    // 存放规则:索引(0-7)表示行,索引对应的值(0-8)表示该行可以存放皇后的列值,以此构建一个类似二位数组
    private static int[] queen = new int[SIZE];
    // 方案数
    private static int step = 0;

    public static void main(String[] args) {
        dfs(0);
    }

    /**
     * @fun 判断当前行可以存放皇后的列
     * @param key : 当前行(可以看作坐标中的 X1)
     * @param value : 当前行对应的列值(可以看作坐标中X1对应的值Y1)
     * @return : 当前行对应的列是否可以存放皇后
     */
    public static boolean isOk(int key, int value) {
        for (int i = 0; i < key; i++) {
            if (i == key || queen[i] == value || (i - key == queen[i] - value) || (key - i == queen[i] - value)) {
                return false;
            }
        }
        return true;
    }

    /**
     * @fun dfs,深度遍历递归函数
     * @param row : 行扫描,问题规模
     */
    public static void dfs(int row) {
        if (row == SIZE) {
            step++;
            print(step);
            return;
        }

        for (int i = 0; i < SIZE; i++) {
            if (isOk(row, i)) {
                queen[row] = i;
                dfs(row + 1);
            }
        }
    }

    /**
     * @fun 打印 8 皇后可行方案
     * @param step : 第几个方案
     */
    public static void print(int step) {
        System.out.println("方案(" + step + "):");
        for (int i = 0; i < SIZE; i++) {
            for (int j = 0; j < SIZE; j++) {
                if (queen[i] == j) { // 皇后位置
                    System.out.print(" @ ");
                } else { // 非皇后
                    System.out.print(" * ");
                }
            }
            System.out.println();
        }
    }

}

五、答案
共有 92 中方案:

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