2013-02-18 39 views
3

这是我在java中的n皇后问题的代码。然而,当它应该是92时,输出为0(在这种情况下为8个皇后的解决方案的数量)。我们应该只使用堆栈和回溯(无递归!!)。我真的被卡住了!任何帮助将不胜感激!在java中使用堆栈和回溯的n皇后拼图

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() 

} 

回答

0

对于初学者来说,conflict()需要检查的s每一个元素,以确定是否存在冲突。

我还没有分析其余的代码,但我也发现conflict()可以悄悄修改s