/** * File : n_queens.c * Created Time: 2023-09-25 * Author : lucas (superrat6@gmail.com) */ #include "../utils/common.h" #define MAX_SIZE 100 /* 回溯算法:n 皇后 */ void backtrack(int row, int n, char state[MAX_SIZE][MAX_SIZE], char ***res, int *resSize, bool cols[MAX_SIZE], bool diags1[2 * MAX_SIZE - 1], bool diags2[2 * MAX_SIZE - 1]) { // 当放置完所有行时,记录解 if (row == n) { res[*resSize] = (char **)malloc(sizeof(char *) * n); for (int i = 0; i < n; ++i) { res[*resSize][i] = (char *)malloc(sizeof(char) * (n + 1)); strcpy(res[*resSize][i], state[i]); } (*resSize)++; 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次对角线 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、次对角线上存在皇后 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, resSize, 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 */ char ***nQueens(int n, int *returnSize) { char state[MAX_SIZE][MAX_SIZE]; // 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_SIZE] = {false}; // 记录列是否有皇后 bool diags1[2 * MAX_SIZE - 1] = {false}; // 记录主对角线上是否有皇后 bool diags2[2 * MAX_SIZE - 1] = {false}; // 记录次对角线上是否有皇后 char ***res = (char ***)malloc(sizeof(char **) * MAX_SIZE); *returnSize = 0; backtrack(0, n, state, res, returnSize, cols, diags1, diags2); return res; } /* Driver Code */ int main() { int n = 4; int returnSize; char ***res = nQueens(n, &returnSize); printf("输入棋盘长宽为%d\n", n); printf("皇后放置方案共有 %d 种\n", returnSize); for (int i = 0; i < returnSize; ++i) { for (int j = 0; j < n; ++j) { printf("["); for (int k = 0; res[i][j][k] != '\0'; ++k) { printf("%c", res[i][j][k]); if (res[i][j][k + 1] != '\0') { printf(", "); } } printf("]\n"); } printf("---------------------\n"); } // 释放内存 for (int i = 0; i < returnSize; ++i) { for (int j = 0; j < n; ++j) { free(res[i][j]); } free(res[i]); } free(res); return 0; }