Ultraviolet A 531: Compromise

Obtain the longest common subsequence (LCS) and print the sequence. Dp Solution: mark the final sequence selection of each node and print it recursively. My problem-solving code is as follows:

 #include <iostream>  #include <cstdio>  #include <cstring>  #include <cstdlib>  #include <algorithm>  using namespace std;    #define max(a,b) (a>b?a:b)  #define WORDLEN 32  #define SENTENCELEN 102  char P1[SENTENCELEN][WORDLEN], P2[SENTENCELEN][WORDLEN];  int dp[SENTENCELEN][SENTENCELEN];  int PathSel[SENTENCELEN][SENTENCELEN];    void PrintSeq(int i, int j)  {      if(i<0 || j<0 || !dp[i][j]) return ;      if(!PathSel[i][j])      {          PrintSeq(i-1,j-1);          if(dp[i][j]==1) cout << P1[i];          else cout << ' ' << P1[i];      }      else if(PathSel[i][j]==1) PrintSeq(i-1,j);      else PrintSeq(i,j-1);     }  int main()  {      int m,n;      while(cin >> P1[0])      {          m = 0, n = 0;          while(P1[m++][0]!='#' && cin >> P1[m]) ; m--;          while(cin >> P2[n] && P2[n++][0]!='#') ; n--;                    dp[0][0] = 0;          for(int i=0; i<m; i++)          {              for(int j=0; j<n; j++)              {                  if(!strcmp(P1[i],P2[j]))                  {                      if(!i || !j) dp[i][j] = 1;                      else dp[i][j] = dp[i-1][j-1]+1;                      PathSel[i][j] = 0;                  }                  else if(!i || !j)                  {                      if(i) { dp[i][j] = dp[i-1][j]; PathSel[i][j] = 1;}                      if(j) { dp[i][j] = dp[i][j-1]; PathSel[i][j] = 2;}                  }                  else if(dp[i-1][j] > dp[i][j-1])                  {                      dp[i][j] = dp[i-1][j];                      PathSel[i][j] = 1;                  }                  else                  {                      dp[i][j] = dp[i][j-1];                      PathSel[i][j] = 2;                  }              }          }          PrintSeq(m-1,n-1);          cout << endl;      }      return 0;  }