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krahets
2023-10-27 20:41:59 +08:00
parent 25ac5de466
commit a42c1d62b9
5 changed files with 34 additions and 33 deletions
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14
docs/chapter_backtracking/n_queens_problem.md
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@ -602,8 +602,8 @@ comments: true
```c title="n_queens.c"
/* 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]) {
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);
@ -635,7 +635,7 @@ comments: true
/* 求解 N 皇后 */
char ***nQueens(int n, int *returnSize) {
char state[MAX_N][MAX_N];
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) {
@ -643,11 +643,11 @@ comments: true
}
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
bool cols[MAX_SIZE] = {false}; // 记录列是否有皇后
bool diags1[2 * MAX_SIZE - 1] = {false}; // 记录主对角线是否有皇后
bool diags2[2 * MAX_SIZE - 1] = {false}; // 记录副对角线是否有皇后
char ***res = (char ***)malloc(sizeof(char **) * MAX_RES);
char ***res = (char ***)malloc(sizeof(char **) * MAX_SIZE);
*returnSize = 0;
backtrack(0, n, state, res, returnSize, cols, diags1, diags2);
return res;