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

 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                 res = res > t1.length() ? res : t1.length();
19         }
20     }
21 }
22 
23 //利用回文子串中心的两边相同
24 void way2()
25 {
26     for (int i = 0; i < str.size(); ++i) {
27         int j;
28         for (j = 1; i - j >= 0 && 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
29         res= max(res, 2 * j - 1);//更新回文子串最大长度
30         for (j = 0; i - j >= 0 && i + j + 1 < str.size() && str[i - j] == str[i + 1 + j]; ++j);//以当前字符为回文中心左侧字符查找最长回文子串
31         res = max(res, 2 * j);//更新回文子串最大长度
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 }