题解：P7995 [USACO21DEC] Walking Home B
一道很普通的计数 $\texttt{dp}$。设 $dp_{i,j,k,l}$ 表示在 $(i,j)$ 位置上，还剩下 $k$ 次行走方向改变，当前的方向为 $l$ 时的方案数。因为只能够向下走，则令 $l \in \{0,1\}$ 表示这两种方向。那么显然从起点开始初始化，对于一个草堆的格子，方案数一定为 $0$，对于其它格子，就会有更改方向或不更改方向的两种转移：
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| for (int l = 0;l <= k;++l)
{
dp[i][j][0][l] += dp[i][j - 1][0][l];
dp[i][j][1][l] += dp[i - 1][j][1][l];
if (l == k) continue;
dp[i][j][0][l] += dp[i][j - 1][1][l + 1];
dp[i][j][1][l] += dp[i - 1][j][0][l + 1];
}
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最后就是答案的统计，根据状态的含义则有 $\sum_{i = 1}^{K} dp_{n,n,i,0} + dp_{n,n,i,1}$。这样就解决了问题，代码如下：
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| #include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <queue>
#define init(x) memset (x,0,sizeof (x))
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
using namespace std;
const int MAX = 55;
const int MOD = 1e9 + 7;
inline int read ();
int t,n,k,ans,dp[MAX][MAX][2][5];
char s[MAX][MAX];
int main ()
{
t = read ();
while (t--)
{
n = read ();k = read ();
ans = 0;
init (dp);
for (int i = 1;i <= n;++i) scanf ("%s",s[i] + 1);
if (s[1][2] != 'H') dp[1][2][0][k] = 1;
if (s[2][1] != 'H') dp[2][1][1][k] = 1;
for (int i = 1;i <= n;++i)
{
for (int j = 1;j <= n;++j)
{
if (i == 1 && j == 1) continue;
if (i == 1 && j == 2) continue;
if (i == 2 && j == 1) continue;
if (s[i][j] == 'H') continue;
for (int l = 0;l <= k;++l)
{
dp[i][j][0][l] += dp[i][j - 1][0][l];
dp[i][j][1][l] += dp[i - 1][j][1][l];
if (l == k) continue;
dp[i][j][0][l] += dp[i][j - 1][1][l + 1];
dp[i][j][1][l] += dp[i - 1][j][0][l + 1];
}
}
}
for (int i = 0;i <= k;++i)
ans += dp[n][n][0][i] + dp[n][n][1][i];
printf ("%d\n",ans);
}
return 0;
}
inline int read ()
{
int s = 0;int f = 1;
char ch = getchar ();
while ((ch < '0' || ch > '9') && ch != EOF)
{
if (ch == '-') f = -1;
ch = getchar ();
}
while (ch >= '0' && ch <= '9')
{
s = s * 10 + ch - '0';
ch = getchar ();
}
return s * f;
}
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