我一直试图用回溯来解决N皇后问题。我在网上找到的大多数方法都涉及向量,这使得我很难将解决方案可视化为互联网上的一些小程序。
我提出的解决方案,给了我很多问题(我有一种感觉与使用的动态2D数组的索引有关)并且我无法使用Dev-C ++调试器解决它。任何帮助和/或建设性的批评受到高度赞赏。提前谢谢了。
以下是我提出的解决方案:
#include<iostream>
#include<string.h>
#include<conio.h>
using namespace std;
void display(char** b, int len);
void initialize(char** &b, int k);
void consider1strow(char ** b, int len);
void markunsafe(char** board, int rowno, int colno);
void marksafe(char** board, int rowno, int colno);
void considerrow(char** board, int rowno);
void backtrack(char** board, int rowno);
bool checksafety(char** board, int rowno, int colno);
void place(char** board, int rowno, int colno);
void solve(char** board, int len);
int state[20] = { 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0 };
int len;
void display(char** board, int len)
{
int i, j;
cout << endl << "The current state of the board:" << endl;
for (i = 0; i < len; i++)
{
for (j = 0; j < len; j++)
{
cout << board[i][j];
}
cout << endl;
}
}
void initialize(char** &b, int k)
{
int i, j;
//create dynamic board
b = new char*[k];
for (i = 0; i < k; i++)
{
b[i] = new char[k];
}
//initialize array
for (i = 0; i < k; i++)
{
for (j = 0; j < k; j++)
{
b[i][j] = '-';
}
}
}
void consider1strow(char ** board, int len)
{
int col;
cout << "Enter the column to try for the first row!";
cin >> col;
board[0][col - 1] = 'Q';
state[0] = col - 1;
markunsafe(board, 0, col - 1);
display(board, len);
}
void markunsafe(char** board, int rowno, int colno)
{
int i, j;
//mark row as unsafe
for (i = 0; i < len; i++)
{
board[rowno][i] = 'x';
}
//mark column as unsafe
for (i = 0; i < len; i++)
{
board[i][colno] = 'x';
}
//mark unsafe diagonals
for (i = 0; i < len; i++)
{
for (j = 0; j < len; j++)
{
if ((rowno + colno) == (i + j))
{
board[i][j] = 'x'; //check if index gives a problem of +/- 1
}
if ((rowno - colno) == (i - j))
{
board[i][j] = 'x'; //check if index gives a problem of +/- 1
}
}
}
board[rowno][colno] = 'Q';
}
void marksafe(char** board, int rowno, int colno)
{
int i, j;
//mark row as safe
for (i = 0; i < len; i++)
{
board[rowno][i] = '-';
}
//mark column as unsafe
for (i = 0; i < len; i++)
{
board[i][colno] = '-';
}
//mark unsafe diagonals
for (i = 0; i < len; i++)
{
for (j = 0; j < len; j++)
{
if ((rowno + colno) == (i + j))
{
board[i][j] = '-'; //check if index gives a problem of +/- 1
}
if ((rowno - colno) == (i - j))
{
board[i][j] = '-'; //check if index gives a problem of +/- 1
}
}
}
}
void considerrow(char** board, int rowno)
{
bool safe = 0;
int i;
for (i = 0; i < len; i++)
{
safe = checksafety(board, rowno, i);
if (safe && (i >= state[rowno]))
{
break;
}
}
if (safe && (i >= state[rowno]))
{
place(board, rowno, i);
}
else if (!safe)
{
backtrack(board, rowno);
}
}
void backtrack(char** board, int rowno)
{
marksafe(board, rowno - 2, state[rowno - 2]);
considerrow(board, rowno);
}
bool checksafety(char** board, int rowno, int colno)
{
if (rowno == 0)
{
return 1;
}
else if (board[rowno][colno] == 'x')
{
return 0;
}
else if (board[rowno][colno] == '-')
{
return 1;
}
}
void place(char** board, int rowno, int colno)
{
board[rowno][colno] = 'Q';
state[rowno] = colno;
markunsafe(board, rowno, colno);
}
void solve(char** board, int len)
{
int i = 0;
if (i == len)
{
display(board, len);
}
else
{
consider1strow(board, len);
for (i = 1; i < len; i++)
{
considerrow(board, i);
}
}
}
int main()
{
char** board;
cout << "Enter the size of the board!";
cin >> len;
initialize(board, len);
solve(board, len);
getch();
}
答案 0 :(得分:1)
它在初始配置后运行,但您不打印它。改变这个(内部解决):
for(i=1;i<len;i++)
{considerrow(board,i);}
为此:
for(i=1; i<len; i++) {
considerrow(board,i);
display(board,len);
}
除此之外,您进行回溯的方式存在问题。如果没有可用的选项,则从上一行中移除后处理(这没关系),然后将正在攻击的每个单元标记为安全(不正常)。问题是这些细胞中的一些可能受到不同女王的攻击,所以你不能将它们标记为安全。此外,您不要在该行上放置不同的女王。我提出了一些解决方案:
首先,使其递归:considerrow
将使用以下行调用自身,如果成功则返回true(1),如果失败则返回false(0)。如果它与下一行失败,你可以使用当前行中的下一个皇后并再次调用considerrow
,直到你成功或用完列,在这种情况下你返回false。
要考虑某一行上的不同女王,您可以做两件事:创建一个董事会副本,您将传递给considerrow
以获取下一行(并因此保留'之前'副本以尝试一个不同的女王),或将每个单元格标记为安全,然后检查所有现有的皇后,以标记不安全的单元格。
编辑:
为了使其递归,我们将使considerrow
使用下一个值调用自身。
bool considerrow(char** board,int rowno) {
//Print the board
display(board,len);
bool safe=0;
int i;
for(i=0; i<len; i++) {
safe=checksafety(board,rowno,i);
if(safe) {
place(board,rowno,i);
//Is this the last row? If so, we suceeded
if (rowno==len-1) return 1;
//Call itself with next row, check if suceeded
if (considerrow(board,rowno+1))
return 1;
else //Failed, try a different row
backtrack(board,rowno);
}
}
return 0; //If we got here, then we ran out of colums. Return failure
}
可以修改回溯功能以恢复当前行,如下所示:
void backtrack(char** board, int rowno) {
//Clear the current row
marksafe(board,rowno,state[rowno]);
//Check that every cell attacked by another queen is marked unsafe
for(int i=0; i<rowno; i++) markunsafe(board,i,state[i]);
}
这样做,解决只需要调用第一行:
void solve(char** board,int len) {
considerrow(board,0);
display(board,len);
}