|
|
@ -222,9 +222,57 @@ comments: true
|
|
|
|
=== "C#"
|
|
|
|
=== "C#"
|
|
|
|
|
|
|
|
|
|
|
|
```csharp title="n_queens.cs"
|
|
|
|
```csharp title="n_queens.cs"
|
|
|
|
[class]{n_queens}-[func]{backtrack}
|
|
|
|
/* 回溯算法:N 皇后 */
|
|
|
|
|
|
|
|
void backtrack(int row, int n, List<List<string>> state, List<List<List<string>>> res,
|
|
|
|
|
|
|
|
bool[] cols, bool[] diags1, bool[] diags2) {
|
|
|
|
|
|
|
|
// 当放置完所有行时,记录解
|
|
|
|
|
|
|
|
if (row == n) {
|
|
|
|
|
|
|
|
List<List<string>> copyState = new List<List<string>>();
|
|
|
|
|
|
|
|
foreach (List<string> sRow in state) {
|
|
|
|
|
|
|
|
copyState.Add(new List<string>(sRow));
|
|
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
res.Add(copyState);
|
|
|
|
|
|
|
|
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 或 主对角线 或 副对角线) 包含皇后
|
|
|
|
|
|
|
|
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;
|
|
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
[class]{n_queens}-[func]{nQueens}
|
|
|
|
/* 求解 N 皇后 */
|
|
|
|
|
|
|
|
List<List<List<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
|
|
|
|
|
|
|
|
List<List<string>> state = new List<List<string>>();
|
|
|
|
|
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
|
|
|
|
|
List<string> row = new List<string>();
|
|
|
|
|
|
|
|
for (int j = 0; j < n; j++) {
|
|
|
|
|
|
|
|
row.Add("#");
|
|
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
state.Add(row);
|
|
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
bool[] cols = new bool[n]; // 记录列是否有皇后
|
|
|
|
|
|
|
|
bool[] diags1 = new bool[2 * n - 1]; // 记录主对角线是否有皇后
|
|
|
|
|
|
|
|
bool[] diags2 = new bool[2 * n - 1]; // 记录副对角线是否有皇后
|
|
|
|
|
|
|
|
List<List<List<string>>> res = new List<List<List<string>>>();
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
backtrack(0, n, state, res, cols, diags1, diags2);
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
return res;
|
|
|
|
|
|
|
|
}
|
|
|
|
```
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
|
|
=== "Swift"
|
|
|
|
=== "Swift"
|
|
|
|