feat: n_queens.c translation (#774)

* feat: tranlating n_queens.c

* Reference to c++ code with additional comments

* Update n_queens.c

* Update n_queens.c

* Format n_queens.c

* Update n_queens.c

keep  the file header information format consistent with others

* fine tune

* fine tune

* Delete mkdocs-en.yml

* Update README.md

---------

Co-authored-by: krahets <krahets@163.com>

codes/c/chapter_backtracking/CMakeLists.txt

@@ -6,3 +6,4 @@ add_executable(preorder_traversal_iii_template preorder_traversal_iii_template.c
 add_executable(subset_sum_i_naive subset_sum_i_naive.c)
 add_executable(subset_sum_i subset_sum_i.c)
 add_executable(subset_sum_ii subset_sum_ii.c)
+add_executable(n_queens n_queens.c)

codes/c/chapter_backtracking/n_queens.c

@@ -0,0 +1,103 @@
+/**
+ * File : n_queens.c
+ * Created Time: 2023-09-25
+ * Author : lucas (superrat6@gmail.com)
+ */
+
+#include "../utils/common.h"
+
+#define MAX_N 100
+#define MAX_RES 1000
+
+struct result {
+    char ***data;
+    int size;
+};
+
+typedef struct result Result;
+
+/* 回溯算法：N 皇后 */
+void backtrack(int row, int n, char state[MAX_N][MAX_N], Result *res, bool cols[MAX_N], bool diags1[2 * MAX_N - 1],
+               bool diags2[2 * MAX_N - 1]) {
+    // 当放置完所有行时，记录解
+    if (row == n) {
+        res->data[res->size] = (char **)malloc(sizeof(char *) * n);
+        for (int i = 0; i < n; ++i) {
+            res->data[res->size][i] = (char *)malloc(sizeof(char) * (n + 1));
+            strcpy(res->data[res->size][i], state[i]);
+        }
+        res->size++;
+        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 */
+Result *nQueens(int n) {
+    char state[MAX_N][MAX_N];
+    // 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
+    for (int i = 0; i < n; ++i) {
+        for (int j = 0; j < n; ++j) {
+            state[i][j] = '#';
+        }
+        state[i][n] = '\0';
+    }
+    bool cols[MAX_N] = {false};           // 记录列是否有皇后
+    bool diags1[2 * MAX_N - 1] = {false}; // 记录主对角线是否有皇后
+    bool diags2[2 * MAX_N - 1] = {false}; // 记录副对角线是否有皇后
+
+    Result *res = malloc(sizeof(Result));
+    res->data = (char ***)malloc(sizeof(char **) * MAX_RES);
+    res->size = 0;
+    backtrack(0, n, state, res, cols, diags1, diags2);
+    return res;
+}
+
+/* Driver Code */
+int main() {
+    int n = 4;
+    Result *res = nQueens(n);
+
+    printf("输入棋盘长宽为%d\n", n);
+    printf("皇后放置方案共有 %d 种\n", res->size);
+    for (int i = 0; i < res->size; ++i) {
+        for (int j = 0; j < n; ++j) {
+            printf("[");
+            for (int k = 0; res->data[i][j][k] != '\0'; ++k) {
+                printf("%c", res->data[i][j][k]);
+                if (res->data[i][j][k + 1] != '\0') {
+                    printf(", ");
+                }
+            }
+            printf("]\n");
+        }
+        printf("---------------------\n");
+    }
+
+    // 释放内存
+    for (int i = 0; i < res->size; ++i) {
+        for (int j = 0; j < n; ++j) {
+            free(res->data[i][j]);
+        }
+        free(res->data[i]);
+    }
+    free(res->data);
+
+    return 0;
+}