Uvs -- 111 History Grading + dp

Uvs -- 111 History Grading + dp

Question:

In fact, it is to find the longest common subsequence of the two sequences.

Ideas:

The input of this question is very difficult. If you understand the input clearly, it will be difficult. The input of a question indicates the position where the number is stored. For example, if the input is 1, 3, and 2, the corresponding sequence of 2 and 4 should be 1, 3, 2, and 4;

The following two pieces of code are provided: one is a classic solution, and the other is the code that I wrote today to convert the problem into the longest path for solving the DAG graph.

The Code is as follows:

#include
 
  #include
  
   #include
   
    using namespace std;int d[30],n,map[30][30];int dp(int i){    if(d[i]>0) return d[i];    d[i]=1;    for(int j=1;j<=n;j++)        if(map[i][j])        {            int t=dp(j)+1;            if(d[i]
    
     

     

     

     
#include
      
       #include
       
        #include
        
         using namespace std;int main(){    int i,j,a[30],b[30],dp[30][30],n;    while(scanf("%d",&n)!=EOF)    {         int x;         for(i=1;i<=n;i++)         {             scanf("%d",&x);             a[x]=i;         }         while(scanf("%d",&x)!=EOF)         {             b[x]=1;             for(i=2;i<=n;i++)             {                 scanf("%d",&x);                 b[x]=i;             }           memset(dp,0,sizeof(dp));           for(i=1;i<=n;i++)              for(j=1;j<=n;j++)              {                     if(a[i]==b[j])                          dp[i][j]=dp[i-1][j-1]+1;                     else                           dp[i][j]=max(dp[i-1][j],dp[i][j-1]);              }              int ans=0;              for(i=1;i<=n;i++)                 for(j=1;j<=n;j++)                      ans=max(ans,dp[i][j]);              printf("%d\n",ans);         }    }  return 0;}