hello-algo/codes/dart/chapter_backtracking/n_queens.dart
liuyuxin 4325974af1
feat: Add Dart codes for chapter_backtracking and chapter_divide_and_conquer (#680)
* feat: Add Dart codes for chapter_backtracking

* feat: Add Dart codes for chapter_divide_and_conquer

* Format Dart Code
2023-08-10 23:48:56 +08:00

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/**
* File: n_queens.dart
* Created Time: 2023-08-10
* Author: liuyuxin (gvenusleo@gmail.com)
*/
/* 回溯算法N 皇后 */
void backtrack(
int row,
int n,
List<List<String>> state,
List<List<List<String>>> res,
List<bool> cols,
List<bool> diags1,
List<bool> diags2,
) {
// 当放置完所有行时,记录解
if (row == n) {
List<List<String>> copyState = [];
for (List<String> sRow in state) {
copyState.add(List.from(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、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] = true;
diags1[diag1] = true;
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] = false;
diags1[diag1] = false;
diags2[diag2] = false;
}
}
}
/* 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 = List.generate(n, (index) => List.filled(n, "#"));
List<bool> cols = List.filled(n, false); // 记录列是否有皇后
List<bool> diags1 = List.filled(2 * n - 1, false); // 记录主对角线是否有皇后
List<bool> diags2 = List.filled(2 * n - 1, false); // 记录副对角线是否有皇后
List<List<List<String>>> res = [];
backtrack(0, n, state, res, cols, diags1, diags2);
return res;
}
/* Driver Code */
void main() {
int n = 4;
List<List<List<String>>> res = nQueens(n);
print("输入棋盘长宽为 $n");
print("皇后放置方案共有 ${res.length}");
for (List<List<String>> state in res) {
print("--------------------");
for (List<String> row in state) {
print(row);
}
}
}