分析：当前点的情况仅由其左边和上边的点决定，然后马会走过的点标记一下即可
1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "string" 5 using namespace std; 6 const int maxn=30; 7 int vis[maxn][maxn]; 8 long long dp[maxn][maxn]; 9 int n,m,x,y; 10 int main() 11 { 12 cin>>n>>m>>x>>y; 13 vis[x][y]=1; 14 vis[x+2][y+1]=1; 15 vis[x+1][y+2]=1; 16 vis[x-1][y+2]=1; 17 vis[x-2][y+1]=1; 18 vis[x-2][y-1]=1; 19 vis[x-1][y-2]=1; 20 vis[x+1][y-2]=1; 21 vis[x+2][y-1]=1; 22 dp[1][0]=1; 23 for(int i=1;i<=n+1;i++){ 24 for(int j=1;j<=m+1;j++){ 25 dp[i][j]=dp[i-1][j]+dp[i][j-1]; 26 if(vis[i-1][j-1]) 27 dp[i][j]=0; 28 } 29 } 30 cout<<dp[n+1][m+1]<<endl; 31 }