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

思路：

dp[i][j]表示s[i]到s[j]所表示的字串是否是回文字串。0表示i-j不是回文串，1代表是。
递推方程：

当s[i] == s[j] : dp[i][j] = dp[i+1][j-1]（i+1）-（j-1）是回文串，那i-j也肯定是了

当s[i] != s[j] : dp[i][j] =0

边界：dp[i][i] = 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

C++：

#include "iostream"
#include "string"
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