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 = 1;
//最普通的遍历
void way1()
{
    for (int i = 0; i < str.length(); ++i)
    {
        for (int j = str.length() - 1; j > i; --j)
        {
            t1.assign(str.begin() + i, str.begin() + j + 1);
            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 = 0; i < str.size(); ++i) {
        int j;
        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
        res= max(res, 2 * j - 1);//更新回文子串最大长度
        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
        res = max(res, 2 * j);//更新回文子串最大长度
    }
}

//使用动态规划
void way3()
{
    int dp[1010][1010];
    for (int i = 0; i < str.length(); i++)
    {
        dp[i][i] = 1;
        if (i < str.length() - 1 && str[i] == str[i + 1])
        {
            dp[i][i + 1] = 1;
            res = 2;
        }
    }
    for (int L = 3; L <= str.length(); L++) {
        for (int i = 0; i + L - 1 < str.length(); i++) {
            int j = i + L - 1;
            if (str[i] == str[j] && dp[i + 1][j - 1] == 1) {
                dp[i][j] = 1;
                res = L;
            }
        }
    }
}

int main()
{
    getline(cin, str);
    way1();
    cout << res << endl;
    return 0;
}