poj 1159 Palindrome

最新推荐文章于 2020-07-18 13:59:15 发布

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本文介绍了一种算法，该算法通过计算给定字符串及其逆序字符串的最长公共子序列来确定最少需要添加多少个字符使原字符串变为回文串。通过动态规划求解，适用于字符串长度最大为5000的情况。

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题目链接：点击打开链接

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

题目大意：给数字符串的长度和字符串s，问要最少添加几个字符才能构成回文串

基本思路：LIS的变形；找出s和其逆序字符串的最长公共字串，然后n－公共字串的长度即使要添加的字母数

<span style="font-size:18px;">///n-最长公共子序列就是要添加的最少字母组成的回文串

#include <iostream>
#include<cstring>

using namespace std;
short int c[5005][5005];

int main()
{
    int n;
    char s[5005];
    char t[5005];
    while(cin>>n)
    {
        cin>>s;
        int top=0;
        for(int i=n-1; i>=0; i--)
        {
            t[top++]=s[i];
        }
        t[top]='\0';
       // cout<<s<<endl;
        //cout<<t<<endl;
        for(int i=0; i<=n; i++)
        {
            c[i][0]=0;
            c[0][i]=0;
        }
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
            {
                if(s[i-1]==t[j-1])
                    c[i][j]=c[i-1][j-1]+1;
                else
                {
                    if(c[i-1][j]>c[i][j-1])
                        c[i][j]=c[i-1][j];
                    else c[i][j]=c[i][j-1];
                }
            }
        }
        cout<<n-c[n][n]<<endl;
    }
    return 0;
}
</span>