45 lines
1.4 KiB
C++
45 lines
1.4 KiB
C++
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#include <iostream>
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using namespace std;
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int dfs(int i, int j, int visited[][10], int cnt, int m, int n) {
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// 终止条件：当遍历到最后一个格子时，如果已经遍历了mn个格子，则符合条件，数量加1
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if (i == m && j == n) {
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if (cnt == m * n)
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return 1;
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else
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return 0;
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}
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// 非终止条件，继续递归
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int res = 0;
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if (i > 1 and !visited[i - 2][j - 1]) {
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visited[i - 2][j - 1] = 1;
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res += dfs(i - 1, j, visited, cnt + 1, m, n);
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visited[i - 2][j - 1] = 0;
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}
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if (j > 1 and !visited[i - 1][j - 2]) {
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visited[i - 1][j - 2] = 1;
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res += dfs(i, j - 1, visited, cnt + 1, m, n);
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visited[i - 1][j - 2] = 0;
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}
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if (i < m and !visited[i][j - 1]) {
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visited[i][j - 1] = 1;
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res += dfs(i + 1, j, visited, cnt + 1, m, n);
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visited[i][j - 1] = 0;
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}
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if (j < n and !visited[i - 1][j]) {
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visited[i - 1][j] = 1;
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res += dfs(i, j + 1, visited, cnt + 1, m, n);
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visited[i - 1][j] = 0;
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}
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return res;
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}
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int main() {
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int m, n;
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cin >> m >> n;
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int visited[m][10] = {0}; // 二维数组表示已经遍历的格子
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visited[0][0] = 1;
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int cnt = 1; // 已经遍历的格子数目
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cout << dfs(1, 1, visited, cnt, m, n) << endl;
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return 0;
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}
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