PKU 1458 Common Subsequence

PKU 1458 Common Subsequence

问题:
http://acm.pku.edu.cn/JudgeOnline/problem?id=1458

思路:
额...最长公共子序列(LCS), 经典动态规划(详情见CLRS)
     if(first[i] == second[j])
         f(i, j) = f(i-1, j-1) + 1
     else
         f(i, j) = max(f(i-1, j), f(i, j-1))

 1  int
 2  dp_lcs()
 3  {
 4       int  i, j;
 5       for (i = 0 ; i <= m; i ++ )
 6          table[i][ 0 ]  =   0 ;
 7       for (j = 0 ; j <= n; j ++ )
 8          table[ 0 ][j]  =   0 ;
 9 
10       for (i = 1 ; i <= m; i ++ ) {
11           for (j = 1 ; j <= n; j ++ ) {
12               if (fst[i - 1 ]  ==  snd[j - 1 ])
13                  table[i][j]  =  table[i - 1 ][j - 1 ]  +   1 ;
14               else
15                  table[i][j]  =  max(table[i - 1 ][j], table[i][j - 1 ]);
16          }
17      }
18       return  table[m][n];
19  }