This is the classic LCS problem. Since it only requires you to print the maximum length, the code can be optimized to use only O(m) space (see here).
My accepted code is as follows.
1 #include <iostream> 2 #include <string> 3 #include <vector> 4 5 using namespace std; 6 7 int lcs(string s, string t) { 8 int m = s.length(), n = t.length(); 9 vector<int> cur(m + 1, 0); 10 for (int j = 1; j <= n; j++) { 11 int pre = 0; 12 for (int i = 1; i <= m; i++) { 13 int temp = cur[i]; 14 cur[i] = (s[i - 1] == t[j - 1] ? pre + 1 : max(cur[i], cur[i - 1])); 15 pre = temp; 16 } 17 } 18 return cur[m]; 19 } 20 21 int main(void) { 22 string s, t; 23 while (getline(cin, s)) { 24 getline(cin, t); 25 printf("%d\n", lcs(s, t)); 26 } 27 return 0; 28 }
Well, try this problem here and get Accepted :)
