poj1159 Palindrome(最長公共子序列)

poj1159 Palindrome(最長公共子序列)

Palindrome Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 52966 Accepted: 18271

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

13368265happystick1159Accepted148K719MSC++720B2014-08-23
 21:42:27
//最少插入字符個數 =字符串長度-正、逆序的最長公共子序列長度
/*
一維數組+short   優化wa了好幾次，無語了 剛開始是以為多組測試數據，一直wa看了半天人家的代碼才知道。。。。就一組測試數據不過這個代碼應該是最優代碼吧 參考的南陽優秀代碼   同樣代碼在hdoj提交wa   無語
Time:2014-8-23 21:47
*/ 
#include
#include
int main(){
	char s1[5005],s2[5005];
	short dp[5005];
	int N;
		scanf("%d",&N);
		scanf("%s",s1);
		for(int i=0;idp[j-1]?temp:dp[j-1];//取dp[i-1][j]和dp[i][j-1]中的較大值 
				}
				old=temp;//上一列的dp[i-1][j]為下一列的dp[i-1][j-1] 
			}
		}
		printf("%d\n",N-dp[N-1]); 
	
return 0;
}