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我想使用堆棧而不是遞歸。我回顧了類似的答案,他們似乎很好,但我真的不明白。我所做的只是使代碼變得更糟。 我想在以下代碼中使用堆棧:http://www.geeksforgeeks.org/backtracking-set-3-n-queen-problem/
下面的代碼通過遞歸實現n皇后問題遞歸,它工作正常。我只需要使用堆棧而不是遞歸。我認爲回溯算法是使用堆棧創建遞歸的好主意。如何使用堆棧而不是遞歸?
任何幫助將不勝感激。
#define N 4
#include<stdio.h>
/* A utility function to print solution */
void printSolution(int board[N][N])
{
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
printf(" %d ", board[i][j]);
printf("\n");
}
}
/* A utility function to check if a queen can be placed on board[row][col]
Note that this function is called when "col" queens are already placeed
in columns from 0 to col -1. So we need to check only left side for
attacking queens */
bool isSafe(int board[N][N], int row, int col)
{
int i, j;
/* Check this row on left side */
for (i = 0; i < col; i++)
{
if (board[row][i])
return false;
}
/* Check upper diagonal on left side */
for (i = row, j = col; i >= 0 && j >= 0; i--, j--)
{
if (board[i][j])
return false;
}
/* Check lower diagonal on left side */
for (i = row, j = col; j >= 0 && i < N; i++, j--)
{
if (board[i][j])
return false;
}
return true;
}
/* A recursive utility function to solve N Queen problem */
bool solveNQUtil(int board[N][N], int col)
{
/* base case: If all queens are placed then return true */
if (col >= N)
return true;
/* Consider this column and try placing this queen in all rows
one by one */
for (int i = 0; i < N; i++)
{
/* Check if queen can be placed on board[i][col] */
if (isSafe(board, i, col))
{
/* Place this queen in board[i][col] */
board[i][col] = 1;
/* recur to place rest of the queens */
if (solveNQUtil(board, col + 1) == true)
return true;
/* If placing queen in board[i][col] doesn't lead to a solution
then remove queen from board[i][col] */
board[i][col] = 0; // BACKTRACK
}
}
/* If queen can not be place in any row in this colum col
then return false */
return false;
}
/* This function solves the N Queen problem using Backtracking. It mainly uses
solveNQUtil() to solve the problem. It returns false if queens cannot be placed,
otherwise return true and prints placement of queens in the form of 1s. Please
note that there may be more than one solutions, this function prints one of the
feasible solutions.*/
bool solveNQ()
{
int board[N][N] = { {0, 0, 0, 0},
{0, 0, 0, 0},
{0, 0, 0, 0},
{0, 0, 0, 0}
};
if (solveNQUtil(board, 0) == false)
{
printf("Solution does not exist");
return false;
}
printSolution(board);
return true;
}
// driver program to test above function
int main()
{
solveNQ();
getchar();
return 0;
}
解決您的問題,您有沒有在這裏告訴我們這個問題。即這個代碼有什麼問題?有什麼症狀?你到目前爲止做了哪些調試? – 2014-12-03 08:26:47
堆棧在哪裏?使用棧而不是遞歸是手動進行遞歸,可以這麼說,當你彈出堆棧的時候,就像堆棧解除一樣。 – ChiefTwoPencils 2014-12-03 08:30:39
@OliverCharlesworth你是對的。我現在編輯它。 – Ege 2014-12-03 08:43:51