解题思路:利用记忆化递归进行搜索
注意事项:注意将终点dp[n][m]设为1
参考代码:
#pragma GCC optimize(1) #pragma GCC optimize(2) #pragma GCC optimize(3,"Ofast","inline") #include<iostream> #define None -1 using namespace std; //记忆化递归 int n,m;//行数列数 int dp[31][31]={0};//dp[i][j]存储从(i,j)到(n,m)的方式数 int path(int i,int j){ if(dp[i][j]!=None) return dp[i][j]; int right_path=0;//向右走后的方式数 int down_path=0;//向下走后的方式数 if(!(i%2==0&&(j+1)%2==0)&&j!=m){ right_path=path(i,j+1); //cout<<"right:"<<"("<<i<<","<<j<<")->("<<i<<","<<j+1<<")"<<endl; } if(!(j%2==0&&(i+1)%2==0)&&i!=n){ down_path=path(i+1,j); //cout<<"down:"<<"("<<i<<","<<j<<")->("<<i+1<<","<<j<<")"<<endl; } dp[i][j]=right_path+down_path; return dp[i][j]; } int main(){ cin>>n>>m; for(int i=1;i<=n;++i){ for(int j=1;j<=m;++j){ dp[i][j]=None; } } dp[n][m]=1; cout<<path(1,1); }
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