java中使用堆栈和回溯的n皇后谜题

Posted 2023-02-21

【中文标题】java中使用堆栈和回溯的n皇后谜题

【英文标题】：n-queens puzzle in java using stacks and backtracking

【发布时间】：2013-02-02 13:27:06

【问题描述】：

这是我在 java 中解决 n-queens 问题的代码。然而，当它应该是 92 时，输出是 0（在这种情况下是 8 个皇后的解的数量）。我们应该只使用堆栈和回溯（没有递归！！）。我真的被困住了！任何帮助将不胜感激！

import java.util.Stack;

public class NQueens 

  //***** fill in your code here *****
  //feel free to add additional methods as necessary

    public static Stack<Integer> s = new Stack<Integer>();
    public static int n ; 
    public static int total; 
    public static int i; 
    public static int Q; 

  //finds and prints out all solutions to the n-queens problem
  public static int solve(int n) 

     // i goes through each row to place a queen
      // x goes through the columns within each row 

      for(int i = 0; i <n; i++) 

          for(int x = 0; x<n; x++)

            if(conflict(x) == false) // loop through each column and checks whether it conflicts with current position of queen

                s.push(x); // no conflict, push x 

                Q = s.get(x); // set current position of queen
                break; //break out of loop to move on to next row, i++ 

                

            else if (conflict(x)==true)
                if(s.isEmpty() == true)
                    break; 
                

                if(x==n-1) // if its looped through all columns, and there's no valid position
                    s.pop(); //pop last entry 
                    i= -1; // and backtrack to previous row, to find another valid position for q in previous row 
                 

            

            if (s.size()==n) // if stack size is n, then stack is full and a solution has been found
                total++; 
                System.out.print(s);// print solution 
                s.pop();
                i= - 1; //backtrack to find next solution
      
            

   
      return total; 
  

public static boolean conflict(int k) 


if (Q==k|| (k-Q)== (i-(i-1))|| (Q-k)== (i-(i-1)) || k == s.pop())  

            return false; //there is a conflict!! k 


return true; //is conflict



  //this method prints out a solution from the current stack
  //(you should not need to modify this method)
  private static void printSolution(Stack<Integer> s) 
    for (int i = 0; i < s.size(); i ++) 
      for (int j = 0; j < s.size(); j ++) 
        if (j == s.get(i))
          System.out.print("Q ");
        else
          System.out.print("* ");
      //for
      System.out.println();
    //for
    System.out.println();  
  //printSolution()

  // ----- the main method -----
  // (you shouldn't need to change this method)
  public static void main(String[] args) 

  int n = 8;

  // pass in parameter n from command line
  if (args.length == 1) 
    n = Integer.parseInt(args[0].trim());
    if (n < 1) 
      System.out.println("Incorrect parameter");
      System.exit(-1);
    //if   
  //if

  int number = solve(n);
  System.out.println("There are " + number + " solutions to the " + n + "-queens problem.");
 //main()

【参考方案1】：

首先，conflict() 需要检查s 的每个元素以确定是否存在冲突。

我没有分析其余的代码，但我也怀疑conflict()可以悄悄地修改s。