Cf-208B-Solitaire-record status DFS

I didn't even think about recording the status during the competition. It's so embarrassing...

Int dp [n] [I] [j] [k]; the length is n, the state of the n-2 is I, the state of the n-1 is j, the status of n is k. Can the operation be successful.

In this way, the time complexity is (52*52*52*52 );

#include
 
  #include#include
  
   #include
   
    #include
    
     #include
     
      #define INF 1000000using namespace std;char va[15]={'2','3','4','5','6','7','8','9','T','J','Q','K','A'};char su[5]={'S','D','H','C'};int dp[60][60][60][60];int a[60];int pan(int x,int y){    if(x%4==y%4)return 1;    if(x/4==y/4)return 1;    return 0;}int dfs(int ns,int x,int y,int z){    if(dp[ns][x][y][z]!=-1)return dp[ns][x][y][z];    if(ns<=3)    {        if(pan(x,z)&&pan(y,z))        {            dp[ns][x][y][z]=1;        }        else dp[ns][x][y][z]=0;        return dp[ns][x][y][z];    }    int leap=0;    if(pan(y,z)&&      dfs(ns-1,a[ns-3],x,z))leap=1;    if(pan(a[ns-3],z)&&dfs(ns-1,z      ,x,y))leap=1;    dp[ns][x][y][z]=leap;    return leap;}int main(){    int n,m,i,j,k;    char str[11];    while(~scanf("%d",&n))    {        memset(dp,-1,sizeof(dp));        for(i=1;i<=n;i++)        {            scanf("%s",str);            for(j=0;j<13;j++)                if(str[0]==va[j])a[i]=j;            for(j=0;j<4;j++)                if(str[1]==su[j])a[i]=a[i]*4+j;        }        if(n<3)        {            if(n==1)cout<<"YES"<