POJ1159——Palindrome

Palindrome

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 53647		Accepted: 18522

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

2

Source

IOI 2000

问你增加多少字符使得字符串变成回文串

我们对字符串和他的逆序串求一个最长公共子序列，这样一来。LCS里的一定是回文的，而其它的一定不会回文，所以必须在对应位置插入那些字符使得他们对称，最后整个串才干够回文

#include <map>    
#include <set>    
#include <list>    
#include <stack>    
#include <queue>    
#include <vector>    
#include <cstdio>    
#include <cmath>    
#include <cstring>    
#include <iostream>    
#include <algorithm>    
    
using namespace std;

char str1[5010];
char str2[5010];
int dp[2][5010];

int main()
{
	int len;
	while (~scanf("%d", &len))
	{
		scanf("%s", str1);
		for (int i = 0; i < len; i++)
		{
			str2[i] = str1[len - i - 1];
		}
		str2[len] = '\0';
		memset (dp, 0, sizeof(dp) );
		for (int i = 1; i <= len; i++)
		{
			for (int j = 1; j <= len; j++)
			{
				if (str1[i - 1] == str2[j - 1])
				{
					dp[i % 2][j] = dp[(i - 1) % 2][j - 1] + 1;
				}
				else
				{
					dp[i % 2][j] = max(dp[(i - 1) % 2][j], dp[i % 2][j - 1]);
				}
			}
		}
		printf("%d\n", len - dp[len % 2][len]);
	}
}