POJ 1159 Palindrome

Palindrome

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 60328 Accepted: 21010

Description

A palindrome(回文) is a symmetrical(匀称的) string, that is, a string read identically(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 the minimal(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 one integer(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 from uppercase(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 considered distinct(明显的).

Output

Your program is to write to standard output(输出). The first line contains one integer(整数), which is the desired minimal(最低的) number.

Sample Input

5Ab3bd

Sample Output

2给你一个字符串，让你计算最少再加进去几个字符，就可以让这个字符串变成回文串。一个最长公共子序列的问题，求出这个字符串的顺序和逆序的最长公共子序列，所求即为字符串的长度减去减去最长公共子序列的长度，因为你只要在不是最长公共子序列的字符在这个字符串的中与这个字符对应的位置再插入一个相同的字符串，他们就对称了。
#include<iostream>#include<cstdio>#include<cstring>using namespace std;char a[5009],b[5009];short dp[5006][5006];int main(){    int n;    while(cin>>n)    {         scanf("%s",a);        memset(dp,0,sizeof(dp));        for(int i=n-1;i>=0;i--) b[n-i-1]=a[i];        for(int i=0;i<n;i++)        for(int j=0;j<n;j++){        if(a[i]==b[j]) dp[i+1][j+1]=dp[i][j]+1;        else dp[i+1][j+1]=dp[i+1][j]>dp[i][j+1]?dp[i+1][j]:dp[i][j+1];        }        cout<<n-dp[n][n]<<endl;        }    return 0;}