PAT Grade --A1040 Longest Symmetric String

Given a string, you are supposed to output the length of the longest symmetric sub-string. For example, given Is PAT&TAP symmetric?, the longest symmetric sub-string is s PAT&TAP s, hence you must output 11.

Input Specification:

Each input file contains one test case which gives a non-empty string of length no more than 1000.

Output Specification:

For each test case, simply print the maximum length in a line.

Sample Input:

Is PAT&TAP symmetric?

Sample Output:

11

 1 #include <iostream>
 2 #include <string>
 3 #include <algorithm>
 4 
 5 using namespace std;
 6 string str, t1, t2;
 7 int res = 1;
 8 //最普通的遍历
 9 void way1()
10 {
11     for (int i = 0; i < str.length(); ++i)
12     {
13         for (int j = str.length() - 1; j > i; --J)
 14          {
 15              t1.assign (str.begin () + I, str.begin () + + J . 1 );
 16              t2.assign (t1.rbegin (), t1.rend ());
 . 17              IF (T1 == T2)
 18 is                  RES = RES> t1.length ()? RES: t1.length ();
 . 19          }
 20 is      }
 21 is  }
 22 is  
23 is  // use both sides of the center of the same palindromic substring 
24  void Way2 ()
 25  {
 26 is      for ( int I = 0 ; I <str.size (); ++ I) {
 27          intJ;
 28          for (J = . 1 ; I - J> = 0 && I + J <str.size () && STR [I + J] == STR [I - J]; J ++); // the current Find the center of the longest character palindromic palindromic substring 
29          RES = max (RES, 2 * J - . 1 ); // update the maximum length of a palindromic substring 
30          for (J = 0 ; I - J> = 0 && I + J + . 1 <str.size () && STR [I - J] == STR [I + . 1 + J]; J ++); // to find the current character is the character to the left center of the longest palindromic palindromic substring 
31          max = RES (RES, 2 * J); // update the maximum length of a palindromic substring 
32      }
 33  }
34 
35 //使用动态规划
36 void way3()
37 {
38     int dp[1010][1010];
39     for (int i = 0; i < str.length(); i++)
40     {
41         dp[i][i] = 1;
42         if (i < str.length() - 1 && str[i] == str[i + 1])
43         {
44             dp[i][i + 1] = 1;
45             res = 2;
46         }
47     }
48     for (int L = 3; L <= str.length(); L++) {
49         for (int i = 0; i + L - 1 < str.length(); i++) {
50             int j = i + L - 1;
51             if (str[i] == str[j] && dp[i + 1][j - 1] == 1) {
52                 dp[i][j] = 1;
53                 res = L;
54             }
55         }
56     }
57 }
58 
59 int main()
60 {
61     getline(cin, str);
62     way1();
63     cout << res << endl;
64     return 0;
65 }