1040 Longest Symmetric String (25分)

1 题目

1040 Longest Symmetric String (25分)
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

2 解析

2.1 题意

求最长回文子串

2.2 思路

dp[i][j]表示S[i]至S[j]所表示的子串是否是回文串，是则为1，不是为0，根据S[i]是否等于S[j],可以把转移情况分为两类，

• 1 若S[i]==S[j]，那么只要S[i+1]至S[j-1]是回文串，S[i]至S[j]就是回文子串；如果S[i+1]至S[j-1]不是回文子串，那么S[i]至S[j]也不是回文子串；
• 2 若S[i] != S[j], 那么S[i]至S[j]一定不是回文子串
  由此得状态转移方程:
  $dp[i][j]=\left\{ \begin{aligned} dp[i+1][j-1],S[i] = S[i] \\ 0,S[i] != S[i] \\ \end{aligned} \right.$

边界：dp[i][i] = 0, dp[i][i+1] = (S[i] == S[i+1])? 1 : 0

3 参考代码

#include <cstdio>
#include <cstring>

const int MAXN = 1010;
int dp[MAXN][MAXN];
char S[MAXN];

int main(int argc, char const *argv[])
{
    fgets(S, MAXN, stdin);
    int len = strlen(S) - 1;
    int ans = 1;

    for (int i = 0; i < len; ++i)
    {
        dp[i][i] = 1;
        if(i < len - 1){
            if(S[i] == S[i + 1]){
                dp[i][i+1] = 1;
                ans = 2;
            }
        }
    }

    for (int L = 3; L <= len; ++L)
    {
        for (int i = 0; i + L - 1 < len; ++i)
        {
            int j = i + L -1;
            if(S[i] == S[j] && dp[i + 1][j - 1] == 1){
                dp[i][j] = 1;
                ans = L;
            }
        }
    }

    printf("%d\n", ans);
    return 0;
}