mirror of
https://github.com/krahets/hello-algo.git
synced 2024-12-27 20:26:29 +08:00
65 lines
2.2 KiB
C++
65 lines
2.2 KiB
C++
/**
|
||
* File: n_queens.cpp
|
||
* Created Time: 2023-05-04
|
||
* Author: Krahets (krahets@163.com)
|
||
*/
|
||
|
||
#include "../utils/common.hpp"
|
||
|
||
/* 回溯算法：N 皇后 */
|
||
void backtrack(int row, int n, vector<vector<string>> &state, vector<vector<vector<string>>> &res, vector<bool> &cols,
|
||
vector<bool> &diags1, vector<bool> &diags2) {
|
||
// 当放置完所有行时，记录解
|
||
if (row == n) {
|
||
res.push_back(state);
|
||
return;
|
||
}
|
||
// 遍历所有列
|
||
for (int col = 0; col < n; col++) {
|
||
// 计算该格子对应的主对角线和副对角线
|
||
int diag1 = row - col + n - 1;
|
||
int diag2 = row + col;
|
||
// 剪枝：不允许该格子所在列、主对角线、副对角线存在皇后
|
||
if (!cols[col] && !diags1[diag1] && !diags2[diag2]) {
|
||
// 尝试：将皇后放置在该格子
|
||
state[row][col] = "Q";
|
||
cols[col] = diags1[diag1] = diags2[diag2] = true;
|
||
// 放置下一行
|
||
backtrack(row + 1, n, state, res, cols, diags1, diags2);
|
||
// 回退：将该格子恢复为空位
|
||
state[row][col] = "#";
|
||
cols[col] = diags1[diag1] = diags2[diag2] = false;
|
||
}
|
||
}
|
||
}
|
||
|
||
/* 求解 N 皇后 */
|
||
vector<vector<vector<string>>> nQueens(int n) {
|
||
// 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
|
||
vector<vector<string>> state(n, vector<string>(n, "#"));
|
||
vector<bool> cols(n, false); // 记录列是否有皇后
|
||
vector<bool> diags1(2 * n - 1, false); // 记录主对角线是否有皇后
|
||
vector<bool> diags2(2 * n - 1, false); // 记录副对角线是否有皇后
|
||
vector<vector<vector<string>>> res;
|
||
|
||
backtrack(0, n, state, res, cols, diags1, diags2);
|
||
|
||
return res;
|
||
}
|
||
|
||
/* Driver Code */
|
||
int main() {
|
||
int n = 4;
|
||
vector<vector<vector<string>>> res = nQueens(n);
|
||
|
||
cout << "输入棋盘长宽为 " << n << endl;
|
||
cout << "皇后放置方案共有 " << res.size() << " 种" << endl;
|
||
for (const vector<vector<string>> &state : res) {
|
||
cout << "--------------------" << endl;
|
||
for (const vector<string> &row : state) {
|
||
printVector(row);
|
||
}
|
||
}
|
||
|
||
return 0;
|
||
}
|