8 Queens Chess Problem

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本文介绍了一种解决经典八皇后问题的方法，通过回溯算法高效地找出所有可能的不冲突棋盘布局方案。该程序避免了穷举所有组合，确保运行效率，并详细展示了输入输出格式及示例。

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8 Queens Chess Problem

In chess it is possible to place eight queens on the board so that no one queen can be taken by any other. Write a program that will determine all such possible arrangements for eight queens given the initial position of one of the queens.

Do not attempt to write a program which evaluates every possible 8 configuration of 8 queens placed on the board. This would require 8⁸evaluations and would bring the system to its knees. There will be a reasonable run time constraint placed on your program.

Input

The first line of the input contains the number of datasets, and it's followed by a blank line. Each dataset will be two numbers separated by a blank. The numbers represent the square on which one of the eight queens must be positioned. A valid square will be represented; it will not be necessary to validate the input.

To standardize our notation, assume that the upper left-most corner of the board is position (1,1). Rows run horizontally and the top row is row 1. Columns are vertical and column 1 is the left-most column. Any reference to a square is by row then column; thus square (4,6) means row 4, column 6.

Each dataset is separated by a blank line.

Output

Output for each dataset will consist of a one-line-per-solution representation.

Each solution will be sequentially numbered $1 \dots N$ . Each solution will consist of 8 numbers. Each of the 8 numbers will be the ROW coordinate for that solution. The column coordinate will be indicated by the order in which the 8 numbers are printed. That is, the first number represents the ROW in which the queen is positioned in column 1; the second number represents the ROW in which the queen is positioned in column 2, and so on.

The sample input below produces 4 solutions. The full 8 $\times$ 8 representation of each solution is shown below.

DO NOT SUBMIT THE BOARD MATRICES AS PART OF YOUR SOLUTION!

   SOLUTION 1           SOLUTION 2           SOLUTION 3           SOLUTION 4

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

Submit only the one-line, 8 digit representation of each solution as described earlier. Solution #1 below indicates that there is a queen at Row 1, Column 1; Row 5, Column 2; Row 8, Column 3; Row 6, Column 4; Row 3,Column 5; ... Row 4, Column 8.

Include the two lines of column headings as shown below in the sample output and print the solutions in lexicographical order.

Print a blank line between datasets.

Sample Input

1

1 1

Sample Output

SOLN       COLUMN
 #      1 2 3 4 5 6 7 8

 1      1 5 8 6 3 7 2 4
 2      1 6 8 3 7 4 2 5
 3      1 7 4 6 8 2 5 3
 4      1 7 5 8 2 4 6 3

#include<iostream>
#include<cstdio>
using namespace std;

int n;
void backtracking( int, int*, int*, int*, int*, int* );
int main()
{
int N;
int row, col, array[8];
bool blank_line;
while( scanf( "%d", &N ) != EOF )
{
blank_line = 0;
for( int i = 0 ; i < N ; i++ )
{
scanf( "%d%d", &row, &col );
row--, col--;
if( blank_line ) printf( "\n" );
printf( "SOLN COLUMN\n" );
printf( " # 1 2 3 4 5 6 7 8\n" );
printf( "\n" );
blank_line = 1;
int rowput[8] = {0}, colput[8] = {0}, leftslash[15] = {0}, rightslash[15] = {0};
rowput[row] = 1;
colput[col] = 1;
leftslash[row+col] = 1;
rightslash[row-col+7] = 1;
n = 0;
array[col] = row;
backtracking( 0, array, rowput, colput, leftslash, rightslash );
}
}
return 0;
}

void backtracking( int i, int array[], int rowput[], int colput[], int leftslash[] , int rightslash[] )
{
if( i == 8 )
{
printf( "%2d ", ++n );
for( int j = 0 ; j < 8 ; j++ )
{
if( j ) printf( " " );
printf( "%d", array[j]+1 );
}
printf( "\n" );
return;
}
if( colput[i] )
{
backtracking( i+1, array, rowput, colput, leftslash, rightslash );
return;
}
for( int j = 0 ; j < 8 ; j++ )
{
if( rowput[j] || leftslash[j+i] || rightslash[j-i+7] )
continue;
rowput[j] = 1;
leftslash[i+j] = 1;
rightslash[j-i+7] = 1;
array[i] = j;
backtracking( i+1, array, rowput, colput, leftslash, rightslash );
rowput[j] = 0;
leftslash[i+j] = 0;
rightslash[j-i+7] = 0;
}
}