PAT甲级——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

 #include <iostream>
 #include <string>
 #include <algorithm>

 using namespace std;
 string str, t1, t2;
 int res = ;
 //最普通的遍历
 void way1()
 {
     for (int i = ; i < str.length(); ++i)
     {
         for (int j = str.length() - ; j > i; --j)
         {
             t1.assign(str.begin() + i, str.begin() + j + );
             t2.assign(t1.rbegin(), t1.rend());
             if (t1 == t2)
                 res = res > t1.length() ? res : t1.length();
         }
     }
 }

 //利用回文子串中心的两边相同
 void way2()
 {
     for (int i = ; i < str.size(); ++i) {
         int j;
         for (j = ; i - j >=  && i + j < str.size() && str[i + j] == str[i - j]; ++j);//以当前字符为回文中心查找最长回文子串
         res= max(res,  * j - );//更新回文子串最大长度
         for (j = ; i - j >=  && i + j +  < str.size() && str[i - j] == str[i +  + j]; ++j);//以当前字符为回文中心左侧字符查找最长回文子串
         res = max(res,  * j);//更新回文子串最大长度
     }
 }

 //使用动态规划
 void way3()
 {
     int dp[][];
     for (int i = ; i < str.length(); i++)
     {
         dp[i][i] = ;
         if (i < str.length() -  && str[i] == str[i + ])
         {
             dp[i][i + ] = ;
             res = ;
         }
     }
     for (int L = ; L <= str.length(); L++) {
         for (int i = ; i + L -  < str.length(); i++) {
             int j = i + L - ;
             if (str[i] == str[j] && dp[i + ][j - ] == ) {
                 dp[i][j] = ;
                 res = L;
             }
         }
     }
 }

 int main()
 {
     getline(cin, str);
     way1();
     cout << res << endl;
     return ;
 }