PAT甲级1040 Longest Symmetric String (25 分)

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

题目大意：

     给出一个字符串，求其最长的回文子串。

 

 

思路：

     用动态规划来解决。

     此处给出简述。

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

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原文：https://blog.csdn.net/liuchuo/article/details/52138902

    

参考代码：

#include
#include
#include
using namespace std;
string s;
int dp[10001][1001];
int main(){
	getline(cin, s);
	int ans = 1;
	for(int i = 0; i < s.size(); ++i){
		dp[i][i] = 1;
		if(i < s.size() - 1 && s[i] == s[i + 1])	dp[i][i + 1] = 1, ans = 2;
	}
	for(int L = 3; L <= s.size(); ++L)
		for(int i = 0; i + L - 1 < s.size(); ++i){
			int j = i + L - 1;
			if(s[i] == s[j] && dp[i + 1][j - 1]) dp[i][j] = 1, ans = L;
		}
	printf("%d\n", ans);
	return 0;
}