N皇后问题|回溯-3

我们已经讨论了骑士之旅和迷宫中的老鼠问题 第一组 和 第二组 分别地让我们讨论N Queen作为另一个可以使用回溯解决的示例问题。 N皇后是在N×N棋盘上放置N个国际象棋皇后的问题，这样就不会有两个皇后互相攻击。例如，下面是4皇后问题的解决方案。

预期输出是一个二进制矩阵，其中皇后所在的块有1s。例如，以下是上述4种皇后解决方案的输出矩阵。

              { 0,  1,  0,  0}
              { 0,  0,  0,  1}
              { 1,  0,  0,  0}
              { 0,  0,  1,  0}

朴素算法 生成板上皇后区的所有可能配置，并打印满足给定约束的配置。

while there are untried configurations
{
   generate the next configuration
   if queens don't attack in this configuration then
   {
      print this configuration;
   }
}

回溯算法 这个想法是从最左边的一列开始，把皇后一个接一个地放在不同的列中。当我们在一列中放置皇后时，我们会检查与已放置皇后的冲突。在当前列中，如果找到没有冲突的行，则将该行和列标记为解决方案的一部分。如果我们没有发现由于冲突导致的此类冲突，那么我们将回溯并返回false。

1) Start in the leftmost column
2) If all queens are placed
    return true
3) Try all rows in the current column. 
   Do following for every tried row.
    a) If the queen can be placed safely in this row 
       then mark this [row, column] as part of the 
       solution and recursively check if placing
       queen here leads to a solution.
    b) If placing the queen in [row, column] leads to
       a solution then return true.
    c) If placing queen doesn't lead to a solution then
       unmark this [row, column] (Backtrack) and go to 
       step (a) to try other rows.
3) If all rows have been tried and nothing worked,
   return false to trigger backtracking.

回溯解决方案的实现

C/C++

/* C/C++ program to solve N Queen Problem using
   backtracking */
#define N 4
#include <stdbool.h>
#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("");
    }
}

/* 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 placed 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 the 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))
                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 the queen cannot be placed in any row in
        this column 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();
    return 0;
}

输出： 1值表示皇后的位置

 0  0  1  0 
 1  0  0  0 
 0  0  0  1 
 0  1  0  0 

is_safe（）函数中的优化 其思想不是检查左右对角线中的每个元素，而是使用对角线的属性： 1.对于每个右对角线，i和j之和是恒定且唯一的，其中i是元素行，j是元素行 元素的列。 2.对于每个左对角线，i和j的差是恒定且唯一的，其中i和j分别是元素的行和列。 回溯解决方案的实施（优化）

C/C++

/* C/C++ program to solve N Queen Problem using
   backtracking */
#define N 4
#include <stdbool.h>
#include <stdio.h>
/* ld is an array where its indices indicate row-col+N-1
 (N-1) is for shifting the difference to store negative 
 indices */
int ld[30] = { 0 };
/* rd is an array where its indices indicate row+col
   and used to check whether a queen can be placed on 
   right diagonal or not*/
int rd[30] = { 0 };
/*column array where its indices indicates column and 
  used to check whether a queen can be placed in that
    row or not*/
int cl[30] = { 0 };
/* 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("");
    }
}

/* 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 the queen can be placed on
          board[i][col] */
        /* A check if a queen can be placed on 
           board[row][col].We just need to check
           ld[row-col+n-1] and rd[row+coln] where
           ld and rd are for left and right 
           diagonal respectively*/
        if ((ld[i - col + N - 1] != 1 &&
                  rd[i + col] != 1) && cl[i] != 1) {
            /* Place this queen in board[i][col] */
            board[i][col] = 1;
            ld[i - col + N - 1] =
                          rd[i + col] = cl[i] = 1;

            /* recur to place rest of the queens */
            if (solveNQUtil(board, col + 1))
                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
            ld[i - col + N - 1] =
                         rd[i + col] = cl[i] = 0;
        }
    }

    /* If the queen cannot be placed in any row in
        this column 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();
    return 0;
}

输出： 1值表示皇后的位置

 0  0  1  0 
 1  0  0  0 
 0  0  0  1 
 0  1  0  0 

打印N-Queen问题的所有解

资料来源： http://see.stanford.edu/materials/icspacs106b/H19-RecBacktrackExamples.pdf http://en.literateprograms.org/Eight_queens_puzzle_%28C%29 http://en.wikipedia.org/wiki/Eight_queens_puzzle 如果您发现任何不正确的地方，或者您想分享有关上述主题的更多信息，请写下评论。