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POJ1159,Palindrome

2024-07-15 19:58:06   223人阅读

Palindrome
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 52543 Accepted: 18113

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


</pre></p><p><pre name="code" class="java">import java.util.Arrays;
import java.util.Scanner;

public class POJ1159_ieayoio {
	static String s;

	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);
		while (cin.hasNext()) {
			int le = cin.nextInt();
			s = cin.next();
			short[][] f = new short[le + 5][le + 5];
			for (int i = 0; i < f.length; i++) {
				Arrays.fill(f[i], (short) 0);
			}
			for (int i = 1; i <= le; i++) {
				f[0][i] = (short) (le - i + 1);
			}
			for (int i = 1; i <= le; i++) {
				f[i][le + 1] = (short) i;
			}
			for (int i = 1; i <= le; i++)
				for (int j = le; j >= i; j--) {
					if (s(i) == s(j))
						f[i][j] = f[i - 1][j + 1];
					else {
						f[i][j] = (short) Math.min(f[i - 1][j] + 1,
								f[i][j + 1] + 1);
					}
				}
			int min = Integer.MAX_VALUE;
			for (int i = 1; i <= le - 1; i++) {
				if (min > f[i][i + 1])
					min = f[i][i + 1];
			}
			for (int i = 1; i <= le; i++) {
				if (min > f[i][i])
					min = f[i][i];
			}
			System.out.println(min);

		}
	}

	static char s(int i) {
		return s.charAt(i - 1);
	}
}


题目大意：求一个字符串最少添加几个字符可以变成回文串


我的思路就是f[i][j]表示左端数第i个字符，左端数在i右端的第j个字符，不过我开始只是觉得和最长公共子序列很像，没想到另一种方法就是求这个字符串和它逆串的最长公共子序列，然后再用总长度减去最长公共子序列的长度即为所求


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