Substring（最长回文串）

最新推荐文章于 2022-04-22 17:01:38 发布

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最新推荐文章于 2022-04-22 17:01:38 发布

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分类专栏： ACM_动态规划往届省赛题

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本文介绍了一种寻找字符串中最长回文子串的方法，并提供了一个C++实现示例。该算法通过比较原始字符串及其反转字符串来找出最长的回文子串，适用于长度在1到50个字符之间的字符串。

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You are given a string input. You are to find the longest substring of input such that the reversal of the substring is also a substring of input. In case of a tie, return the string that occurs earliest in input.

Note well: The substring and its reversal may overlap partially or completely. The entire original string is itself a valid substring . The best we can do is find a one character substring, so we implement the tie-breaker rule of taking the earliest one first.

Input

The first line of input gives a single integer, 1 ≤ N ≤ 10, the number of test cases. Then follow, for each test case, a line containing between 1 and 50 characters, inclusive. Each character of input will be an uppercase letter ('A'-'Z').

Output

Output for each test case the longest substring of input such that the reversal of the substring is also a substring of input

Sample Input

3                   
ABCABA
XYZ
XCVCX

Sample Output

ABA
X
XCVCX

AC代码：

#include<bits/stdc++.h>
using namespace std;
string s,ss;
int i,j,k,l;
int N;
int dp[55][55];
void sss(){
	ss="";
	for(i=s.size();i>-1;i--){
		ss+=s[i];
	}	
}
int main()
{
	while(cin>>N){
		while(N--){
			cin>>s;
			memset(dp,0,sizeof(dp));
		    sss();
			k=l=0;
			for(i=1;i<=s.size();i++)
            for(j=1;j<=ss.size();j++){
                if(s[i-1]==ss[j-1]){
                    dp[i][j]=dp[i-1][j-1]+1;
                }
                if(l<dp[i][j]){
                    l=dp[i][j];  //更新 
                    k=i-1;
                }
            }
            for(i=k-l+1;i<=k;i++)
			cout<<s[i];
			cout<<endl; 
		}
	}
	return 0;
}