nyoj308 Substring(第四届河南省程序设计大赛)

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Substring

1000 ms  |  内存限制： 65535

1

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.

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'). 输出 Output for each test case the longest substring of input such that the reversal of the substring is also a substring of input 样例输入

3                   
ABCABA
XYZ
XCVCX

样例输出

ABA
X
XCVCX

来源 ​​第四届河南省程序设计大赛​​

上传者 ​​张云聪​​

我只想说这不是回文串。。

一个测试数组abvcba 答案ab

首先把原字符串翻转

把该题理解为最长公共子串，使用动态规划可以解决。

import java.util.Scanner;


public class Main {
  public static void main(String[] args) {
    Scanner sca=new Scanner(System.in);
    int n=sca.nextInt();
    while(n-->0){
      int dp[][]=new int[55][55];
      String a=sca.next();
      int len=a.length();
      String b=new StringBuffer(a).reverse().toString();
      int res=0;
      int index=0;
      for(int i=1;i<=len;i++){
        for(int j=1;j<=len;j++){
          if(a.charAt(i-1)==b.charAt(j-1)){
            dp[i][j]=dp[i-1][j-1]+1;
          }
          if(dp[i][j]>res){
            res=dp[i][j];
            index=i;
          }
        }
      }
      System.out.println(a.substring(index-res, index));
    }
  }

}