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/**
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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_SIZE 100
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/* 回溯算法：n 皇后 */
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void backtrack(int row, int n, char state[MAX_SIZE][MAX_SIZE], char ***res, int *resSize, bool cols[MAX_SIZE],
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bool diags1[2 * MAX_SIZE - 1], bool diags2[2 * MAX_SIZE - 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[*resSize] = (char **)malloc(sizeof(char *) * n);
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for (int i = 0; i < n; ++i) {
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res[*resSize][i] = (char *)malloc(sizeof(char) * (n + 1));
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strcpy(res[*resSize][i], state[i]);
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}
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(*resSize)++;
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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次对角线
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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、次对角线上存在皇后
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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, resSize, 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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char ***nQueens(int n, int *returnSize) {
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char state[MAX_SIZE][MAX_SIZE];
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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_SIZE] = {false}; // 记录列是否有皇后
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bool diags1[2 * MAX_SIZE - 1] = {false}; // 记录主对角线上是否有皇后
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bool diags2[2 * MAX_SIZE - 1] = {false}; // 记录次对角线上是否有皇后
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char ***res = (char ***)malloc(sizeof(char **) * MAX_SIZE);
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*returnSize = 0;
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backtrack(0, n, state, res, returnSize, 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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int returnSize;
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char ***res = nQueens(n, &returnSize);
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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", returnSize);
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for (int i = 0; i < returnSize; ++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[i][j][k] != '\0'; ++k) {
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printf("%c", res[i][j][k]);
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if (res[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 < returnSize; ++i) {
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for (int j = 0; j < n; ++j) {
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free(res[i][j]);
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}
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free(res[i]);
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}
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free(res);
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return 0;
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}
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