POJ1159：Palindrome（LCS）

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本文介绍了一种算法，用于确定将任意给定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与其反转字符串之间的最长公共子序列来得出答案。

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Palindrome

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 62132		Accepted: 21652

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

Source

IOI 2000

题意：给一个字符串，求最少添加几个字符使其变成回文串。

思路：将字符串倒置形成新的字符串，求两者的最长公共子序列ans，字符串长度-ans即为答案。

# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
# define MAXN 5000
using namespace std;
int main()
{
    int dp[2][MAXN+1], len;//dp数组压缩内存。
    char a[MAXN+3], b[MAXN+3];
    while(~scanf("%d",&len))
    {
        scanf("%s",a+1);
        memset(dp, 0, sizeof(dp));
        for(int i=1; i<=len; ++i)
            b[i] = a[len-i+1];
        int t=1;
        for(int i=1; i<=len; ++i)
        {
            t = 1-t;
            for(int j=1; j<=len; ++j)
            {
                if(a[i]==b[j])
                    dp[t][j] = dp[1-t][j-1] + 1;
                else
                    dp[t][j] = max(dp[t][j-1], dp[1-t][j]);
            }
        }
        printf("%d\n",len-dp[t][len]);
    }
    return 0;
}