AcWing 3502. 不同路径数
原题链接
简单
作者:
回归线
,
2021-05-14 18:57:01
,
所有人可见
,
阅读 331
#include <iostream>
#include <unordered_set>
using namespace std;
const int N = 5;
int a[N][N];
unordered_set<int> S;
int m, n, k;
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
void dfs(int i, int j, int step, int num) {
if (step == k) {
S.insert(num);
return;
}
for (int d = 0; d < 4; ++d) {
int x = i + dx[d];
int y = j + dy[d];
if (x >= 0 && x < n && y >= 0 && y < m) {
dfs(x, y, step + 1, num * 10 + a[x][y]);
}
}
}
int main() {
cin >> n >> m >> k;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
cin >> a[i][j];
}
}
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
dfs(i, j, 0, a[i][j]);
}
}
cout << S.size() << endl;
return 0;
}