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* Revised the book * Update the book with the second revised edition * Revise base on the manuscript of the first edition
65 lines
2.2 KiB
C++
65 lines
2.2 KiB
C++
/**
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* File: n_queens.cpp
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* Created Time: 2023-05-04
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* Author: Krahets (krahets@163.com)
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*/
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#include "../utils/common.hpp"
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/* 回溯算法：N 皇后 */
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void backtrack(int row, int n, vector<vector<string>> &state, vector<vector<vector<string>>> &res, vector<bool> &cols,
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vector<bool> &diags1, vector<bool> &diags2) {
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// 当放置完所有行时，记录解
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if (row == n) {
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res.push_back(state);
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return;
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}
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// 遍历所有列
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for (int col = 0; col < n; col++) {
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// 计算该格子对应的主对角线和次对角线
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int diag1 = row - col + n - 1;
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int diag2 = row + col;
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// 剪枝：不允许该格子所在列、主对角线、次对角线上存在皇后
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if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
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// 尝试：将皇后放置在该格子
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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：将该格子恢复为空位
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state[row][col] = "#";
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cols[col] = diags1[diag1] = diags2[diag2] = false;
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}
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}
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}
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/* 求解 N 皇后 */
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vector<vector<vector<string>>> nQueens(int n) {
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// 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
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vector<vector<string>> state(n, vector<string>(n, "#"));
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vector<bool> cols(n, false); // 记录列是否有皇后
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vector<bool> diags1(2 * n - 1, false); // 记录主对角线上是否有皇后
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vector<bool> diags2(2 * n - 1, false); // 记录次对角线上是否有皇后
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vector<vector<vector<string>>> res;
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backtrack(0, n, state, res, cols, diags1, diags2);
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return res;
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}
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/* Driver Code */
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int main() {
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int n = 4;
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vector<vector<vector<string>>> res = nQueens(n);
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cout << "输入棋盘长宽为 " << n << endl;
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cout << "皇后放置方案共有 " << res.size() << " 种" << endl;
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for (const vector<vector<string>> &state : res) {
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cout << "--------------------" << endl;
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for (const vector<string> &row : state) {
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printVector(row);
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}
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}
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return 0;
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}
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