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=begin
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File: n_queens.rb
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Created Time: 2024-05-21
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Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
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=end
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### 回溯演算法：n 皇后 ###
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def backtrack(row, n, state, res, cols, diags1, diags2)
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# 當放置完所有行時，記錄解
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if row == n
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res << state.map { |row| row.dup }
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return
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end
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# 走訪所有列
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for col in 0...n
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# 計算該格子對應的主對角線和次對角線
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diag1 = row - col + n - 1
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diag2 = row + col
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# 剪枝：不允許該格子所在列、主對角線、次對角線上存在皇后
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if !cols[col] && !diags1[diag1] && !diags2[diag2]
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# 嘗試：將皇后放置在該格子
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state[row][col] = "Q"
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cols[col] = diags1[diag1] = diags2[diag2] = true
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# 放置下一行
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backtrack(row + 1, n, state, res, cols, diags1, diags2)
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# 回退：將該格子恢復為空位
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state[row][col] = "#"
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cols[col] = diags1[diag1] = diags2[diag2] = false
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end
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end
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end
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### 求解 n 皇后 ###
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def n_queens(n)
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# 初始化 n*n 大小的棋盤，其中 'Q' 代表皇后，'#' 代表空位
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state = Array.new(n) { Array.new(n, "#") }
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cols = Array.new(n, false) # 記錄列是否有皇后
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diags1 = Array.new(2 * n - 1, false) # 記錄主對角線上是否有皇后
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diags2 = Array.new(2 * n - 1, false) # 記錄次對角線上是否有皇后
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res = []
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backtrack(0, n, state, res, cols, diags1, diags2)
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res
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end
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### Driver Code ###
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if __FILE__ == $0
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n = 4
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res = n_queens(n)
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puts "輸入棋盤長寬為 #{n}"
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puts "皇后放置方案共有 #{res.length} 種"
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for state in res
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puts "--------------------"
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for row in state
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p row
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end
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end
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end
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