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* 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>
103 lines
3 KiB
C
103 lines
3 KiB
C
/**
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* File : n_queens.c
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* Created Time: 2023-09-25
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* Author : lucas (superrat6@gmail.com)
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*/
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#include "../utils/common.h"
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#define MAX_N 100
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#define MAX_RES 1000
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struct result {
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char ***data;
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int size;
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};
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typedef struct result Result;
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/* 回溯算法：N 皇后 */
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void backtrack(int row, int n, char state[MAX_N][MAX_N], Result *res, bool cols[MAX_N], bool diags1[2 * MAX_N - 1],
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bool diags2[2 * MAX_N - 1]) {
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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->data[res->size] = (char **)malloc(sizeof(char *) * n);
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for (int i = 0; i < n; ++i) {
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res->data[res->size][i] = (char *)malloc(sizeof(char) * (n + 1));
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strcpy(res->data[res->size][i], state[i]);
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}
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res->size++;
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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副对角线
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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、副对角线存在皇后
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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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Result *nQueens(int n) {
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char state[MAX_N][MAX_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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for (int i = 0; i < n; ++i) {
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for (int j = 0; j < n; ++j) {
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state[i][j] = '#';
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}
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state[i][n] = '\0';
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}
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bool cols[MAX_N] = {false}; // 记录列是否有皇后
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bool diags1[2 * MAX_N - 1] = {false}; // 记录主对角线是否有皇后
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bool diags2[2 * MAX_N - 1] = {false}; // 记录副对角线是否有皇后
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Result *res = malloc(sizeof(Result));
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res->data = (char ***)malloc(sizeof(char **) * MAX_RES);
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res->size = 0;
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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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Result *res = nQueens(n);
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printf("输入棋盘长宽为%d\n", n);
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printf("皇后放置方案共有 %d 种\n", res->size);
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for (int i = 0; i < res->size; ++i) {
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for (int j = 0; j < n; ++j) {
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printf("[");
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for (int k = 0; res->data[i][j][k] != '\0'; ++k) {
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printf("%c", res->data[i][j][k]);
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if (res->data[i][j][k + 1] != '\0') {
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printf(", ");
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}
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}
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printf("]\n");
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}
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printf("---------------------\n");
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}
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// 释放内存
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for (int i = 0; i < res->size; ++i) {
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for (int j = 0; j < n; ++j) {
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free(res->data[i][j]);
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}
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free(res->data[i]);
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}
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free(res->data);
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return 0;
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}
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