Palindrome

C - Palindrome

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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

5Ab3bd

Sample Output

2

找最多添加多少个字母变成回文串。 用字符串本身与它的逆序串求一下公共子序列（LCS）就好了。求完用字符串总长减去LCS的长度就是了.

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;const int maxn = 5010;char s[maxn],str[maxn];int dp[2][maxn];int n;int main(){while(scanf("%d",&n) != EOF){memset(dp,0,sizeof(dp));scanf("%s",str);for(int i=0; i<n; ++i){s[n-1-i] = str[i];}for(int i=1; i<=n; ++i){for(int j=1; j<=n; j++){if(s[i-1] == str[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",n - dp[n%2][n]);}return 0;}