1040. Longest Symmetric String (25)-PAT甲级真题

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1040. Longest Symmetric String (25)
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

分析：

dp[i][j]表示s[i]到s[j]所表示的字串是否是回文字串。只有0和1
递推方程：
- 当s[i] == s[j] : dp[i][j] = dp[i+1][j-1]
- 当s[i] != s[j] : dp[i][j] =0
- 边界：dp[i][j] = 1, dp[i][i+1] = (s[i] == s[i+1]) ? 1 : 0
因为i、j如果从小到大的顺序来枚举的话，无法保证更新dp[i][j]的时候dp[i+1][j-1]已经被计算过。因此不妨考虑按照字串的长度和子串的初试位置进行枚举，即第一遍将长度为3的子串的dp的值全部求出，第二遍通过第一遍结果计算出长度为4的子串的dp的值...这样就可以避免状态无法转移的问题
首先初始化dp[i][i] = 1, dp[i][i+1]，把长度为1和2的都初始化好，然后从L = 3开始一直到 L <= len 根据动态规划的递归方程来判断

#include <iostream>using namespace std;int dp[1010][1010];int main() {    string s;    getline(cin, s);    int len = s.length(), ans = 1;    for(int i = 0; i < len; i++) {        dp[i][i] = 1;        if(i < len - 1 && 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", ans);    return 0;}

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