poj 1159Palindrome(dp lcs变形)_大写字母dp的变形字体-优快云博客

本文链接：https://blog.youkuaiyun.com/SDUTyangkun/article/details/52231702

本文介绍了一种算法，用于确定将任意字符串转换为回文串所需的最少字符插入数量。通过寻找原始字符串中最长的回文子序列，该算法有效地解决了问题。

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Palindrome

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 60328		Accepted: 21010

Description

A palindrome(回文) is asymmetrical(匀称的) string, that is, a string readidentically(同一地) from left to right as well as from right to left. You are to write a program which, given a string, determines theminimal(最低的) 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(改变) 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(投入). The first line contains oneinteger(整数): the length of the input string N, 3 <= N <= 5000. The second line contains one string with length N. The string is formed fromuppercase(以大写字母印刷) letters from 'A' to 'Z',lowercase(小写字母) letters from 'a' to 'z' anddigits(数字) from '0' to '9'. Uppercase and lowercase letters are to be considereddistinct(明显的).

Output

Your program is to write to standard output(输出). The first line contains oneinteger(整数), which is the desiredminimal(最低的) number.

Sample Input

5
Ab3bd

Sample Output

2
//用int MLE  改成short就ac了  添加最少字母 其实就是 在给定的字符串里里面找最长的回文串  然后剪掉最长的剩下的就是你要最少补充的了~~~~
最长回文 就是正着读和到着读 最长的公共子序列  恩  就这么简单
#include<stdio.h>
#include<string.h>
#define max(a,b) a>b?a:b
short dp[5001][5001];
char x[5000],y[5000];
int main()
{
    int n;
    int i,j,k;

    while(~scanf("%d",&n))
    {
        scanf("%s",x);
        for(i=n-1,j=0;i>=0;i--)
        y[j++]=x[i];

        dp[0][0]=0;
        for(i=1;i<=n;i++)
        {
            dp[i][0]=0;
            dp[0][i]=0;
        }

        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
        {
            if(x[i-1]==y[j-1])
                dp[i][j]=dp[i-1][j-1]+1;
            else
                dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
        }
        printf("%d\n",n-dp[n][n]);

    }
}