hello-algo/codes/c/chapter_backtracking/n_queens.c
krahets 1e49574332 Simplify struct declarations of C.
Use PascalCase for all structs in C.
SImplify n_queens.c
Format C code for chapter of graph.
2023-10-18 02:16:26 +08:00

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/**
* 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
/* 回溯算法N 皇后 */
void backtrack(int row, int n, char state[MAX_N][MAX_N], char ***res, int *resSize, 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[*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, 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_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
char ***res = (char ***)malloc(sizeof(char **) * MAX_RES);
*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;
}