Palindrome - POJ 1159 dp-优快云博客

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/37695111

本文探讨了一个算法问题，即如何通过最少的字符插入操作，将任意给定字符串转换为回文串。详细介绍了使用动态规划解决此问题的方法，并提供了相应的AC代码。

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Palindrome

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 51741		Accepted: 17811

Description

A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are to write a program which, given a string, determines the minimal number of characters to be inserted into the string in order to obtain a palindrome.

As an example, by inserting 2 characters, the string "Ab3bd" can be transformed into a palindrome ("dAb3bAd" or "Adb3bdA"). However, inserting fewer than 2 characters does not produce a palindrome.

Input

Your program is to read from standard input. The first line contains one integer: the length of the input string N, 3 <= N <= 5000. The second line contains one string with length N. The string is formed from uppercase letters from 'A' to 'Z', lowercase letters from 'a' to 'z' and digits from '0' to '9'. Uppercase and lowercase letters are to be considered distinct.

Output

Your program is to write to standard output. The first line contains one integer, which is the desired minimal number.

Sample Input

5
Ab3bd

Sample Output

题意：问最少需要补几个字符，使得其成为一个回文串。

思路：最简单的dp了，具体见代码吧。一开始还在考虑是挂在dp里，还是挂在水题里，想了想，简单的dp也是dp，别拿村长不当干部。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
short dp[5010][5010];
char s[5010];
int main()
{ int n,i,j,k;
  scanf("%d",&n);
  scanf("%s",s);
  for(i=n;i>=1;i--)
   for(j=i+1;j<=n;j++)
   { dp[i][j]=min(dp[i+1][j]+1,dp[i][j-1]+1);
     if(s[i-1]==s[j-1])
      dp[i][j]=min(dp[i][j],dp[i+1][j-1]);
    //printf("%d %d %d\n",i,j,dp[i][j]);
   }
  printf("%d\n",dp[1][n]);
}