[2016-05-09] [51nod] [1006 longest common subsequence LCS]

• Time: 2016-05-09-21:12:54
• Title Number: [2016-05-09][51nod][1006 longest common sub-sequence LCS]
• Main topic: [2016-05-09][51nod][1006 longest common sub-sequence lcs].md
• Analysis: Dynamic Planning
  • DP[I][J] Represents the length of the longest common subsequence of string A in the first position, string B at position J
  • DP[I][J] = dp[i-1][j-1] + 1 if a[i] = = A[j]
  • else dp[i][j] = = Max (Dp[i-1][j], dp[i][j-1]);
  • The maximum length is dp[n][m], n is the length of a, m is the length of B
  • To restore a string, just go back to the place where the dp[i][j] just started to change.

  
 

   
  

  
#include<stdio.h>
   
  

  
#include<string.h>
   
  

  
#include<algorithm>
   
  

  
using namespace std;
   
  

  
const int maxn = 1E3 + 10;
   
  

  
char a[maxn],b[maxn],ans[maxn];
   
  

  
int dp[maxn][maxn];
   
  

  
int main(){
   
  

  
 scanf   ( "%s%s " ,  a  +    1   b  +    1  );   
   
  

  
 int n = strlen(a+1),m = strlen(b+1); 
   
  

  
 memset(dp,0,sizeof(dp)); 
   
  

  
  for   ( int   i  =    1    ;   I  <=   n     ++  i  ) {  
   
  

  
 for(int j = 1 ; j <= m ; ++j){
   
  

  
 if(a[i] == b[j]){
   
  

  
 DP  [ i  ][ j      =   DP  [ i  - 1  ][ j  - 1      +    1    
   
  

  
 }else dp[i][j] = max(dp[i][j-1],dp[i-1][j]);
   
  

  
 }
   
  

  
 } 
   
  

  
 int cur = 0;
   
  

  
  for   ( int   i  =   n   j  =   m   DP  [ i  ][ j  ];-- i  ,-- j   //return to the place where the value was first updated   
   
  

  
  while   ( DP  [ i  ][ j  ]    ==   DP  [ i  -   1  ][ j     -- i    
   
  

  
  while   ( DP  [ i  ][ j  ]    ==   DP  [ i  ][ j  -   1  ])    -- j  ;   
   
  

  
 ans[cur++] = a[i];
   
  

  
 }
   
  

  
 reverse(ans,ans+cur);
   
  

  
 ans[cur] = ‘\0‘;
   
  

  
 printf("%s\n",ans);
   
  

  
 return 0;
   
  

  
}
  
 

From for notes (Wiz)

[2016-05-09][51nod][1006 longest common subsequence LCS]