PAT甲级——A1040 Longest Symmetric String
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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
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