链接
代码1 动态规划
f [ i ] [ j ] [ k ] [ s t ] f[i][j][k][st] f[i][j][k][st] 表示当前在 ( i , j ) (i,j) (i,j) 位置上,且转向次数为 k k k 次,且上一步是左边的点( s t = 0 st=0 st=0)或上边的点( s t = 1 st=1 st=1)的所有方案的集合,存方案数量。
// #pragma GCC optimize(3)
#include <set>
#include <map>
#include <cmath>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <cstdio>
#include <bitset>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_set>
#include <unordered_map>
#define endl '\n'
#define fi first
#define se second
#define PI acos(-1)
#define LL long long
#define INF 0x3f3f3f3f
#define lowbit(x) (-x&x)
#define PII pair<int, int>
#define ULL unsigned long long
#define PIL pair<int, long long>
#define mem(a, b) memset(a, b, sizeof a)
#define rev(x) reverse(x.begin(), x.end())
#define IOS ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
const int N = 60;
int n, K;
char g[N][N];
int dp[N][N][5][2];
void solve() {
int tt;
cin >> tt;
while (tt -- ) {
mem(dp, 0);
cin >> n >> K;
for (int i = 1; i <= n; i ++ ) cin >> g[i] + 1;
dp[1][1][0][0] = dp[1][1][0][1] = 1;
for (int i = 1; i <= n; i ++ ) {
for (int j = 1; j <= n; j ++ ) {
if (i == 1 && j == 1) continue;
if (g[i][j] == 'H') continue;
for (int k = 0; k <= K; k ++ ) {
if ((i == 1 || j == 1) && k > 0) continue;
dp[i][j][k][0] = dp[i - 1][j][k][0];
dp[i][j][k][1] = dp[i][j - 1][k][1];
if (k > 0) {
dp[i][j][k][0] += dp[i - 1][j][k - 1][1];
dp[i][j][k][1] += dp[i][j - 1][k - 1][0];
}
}
}
}
int ans = 0;
for (int k = 0; k <= K; k ++ ) {
ans += dp[n][n][k][0] + dp[n][n][k][1];
}
cout << ans << endl;
}
}
int main() {
IOS;
solve();
return 0;
}
代码2 DFS(O3优化)
#pragma GCC optimize(3)
#include <set>
#include <map>
#include <cmath>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <cstdio>
#include <bitset>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_set>
#include <unordered_map>
#define endl '\n'
#define fi first
#define se second
#define PI acos(-1)
#define LL long long
#define INF 0x3f3f3f3f
#define lowbit(x) (-x&x)
#define PII pair<int, int>
#define ULL unsigned long long
#define PIL pair<int, long long>
#define mem(a, b) memset(a, b, sizeof a)
#define rev(x) reverse(x.begin(), x.end())
#define IOS ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
const int N = 100;
int n, k, ans;
char g[N][N];
int dx[2] = {0, 1}, dy[2] = {1, 0};
/* x,y--当前坐标 last--上一次的方向 sum--转向次数 */
void dfs(int x, int y, int last, int sum) {
if (sum > k) return;
if (x == n && y == n) {
ans ++ ;
return;
}
for (int i = 0; i < 2; i ++ ) {
int xx = x + dx[i], yy = y + dy[i];
if (xx > n || yy > n) continue;
if (g[xx][yy] == 'H') continue;
dfs(xx, yy, i, sum + (last == -1 ? 0 : (last != i)));
}
}
void solve() {
int tt;
cin >> tt;
// tt = 1;
while (tt -- ) {
mem(g, 0);
ans = 0;
cin >> n >> k;
for (int i = 1; i <= n; i ++ ) cin >> g[i] + 1;
dfs(1, 1, -1, 0);
cout << ans << endl;
}
}
int main() {
IOS;
solve();
return 0;
}