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@@ -601,25 +601,17 @@ comments: true
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=== "C"
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```c title="n_queens.c"
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/* 放置结果 */
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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,
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bool cols[MAX_N], bool diags1[2 * MAX_N - 1], bool diags2[2 * MAX_N - 1]) {
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void backtrack(int row, int n, char state[MAX_N][MAX_N], char ***res, int *resSize, bool cols[MAX_N],
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bool diags1[2 * MAX_N - 1], 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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res[*resSize] = (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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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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res->size++;
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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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@@ -633,7 +625,7 @@ comments: true
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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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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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@@ -642,7 +634,7 @@ comments: true
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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 ***nQueens(int n, int *returnSize) {
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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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@@ -655,10 +647,9 @@ comments: true
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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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char ***res = (char ***)malloc(sizeof(char **) * MAX_RES);
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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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```
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