POJ 1159 Palindrome【LCS+滚动数组】【水题】

Palindrome

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 63833		Accepted: 22254

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

给你一个字符串，可在任意位置添加字符，最少再添加几个字符，可以使这个字符串成为回文字符串。
设原序列S的逆序列为S' ，则最少需要补充的字母数 = 原序列S的长度 —  S和S'的最长公共子序列长度

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <string>
#include <map>
#include <cstring>
#define INF 0x3f3f3f3f
#define ms(x,y) memset(x,y,sizeof(x))
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;

const int maxn = 5010;
const int mod = 998244353;

int dp[2][maxn];

int main()
{
	string a, b;
	int n;
	cin >> n;
	cin >> a;
	b = a;
	reverse(a.begin(), a.end());
	int ans = -INF;
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			if (a[i] == b[j])
				dp[i % 2][j] = dp[(i + 1) % 2][j - 1] + 1;
			else
				dp[i % 2][j] = max(dp[i % 2][j - 1], dp[(i + 1) % 2][j]);
			ans = max(ans, dp[i % 2][j]);
		}
	}
	cout << n - ans;
	return 0;
}