PKUOJ1159 Palindrome

 

Palindrome

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 31933 Accepted: 10666

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

【思路】求字符串A和A的逆序字符串A*的LCS(A,A*)   结果是len-LCS。出现过MLE，后将状态数组改为short的动态数组，险过！

#include "stdafx.h"#include<fstream>#include <algorithm>#include<vector>#include<map>#include<set>#include<string>#include<iostream>using namespace std;#define max(x,y) ((x)>(y)?(x):(y))short** rst;inline int LCS(string &s1,string &s2){ int i,j; int len1 = s1.length(); rst = new short*[len1+2]; for(i=0;i<len1+2;i++) {  rst[i] = new short[len1+2];  for(j=0;j<len1+2;j++)   rst[i][j] = 0; }  for(i=0;i<len1;i++)  for(j=0;j<len1;j++)   if(s1[i] == s2[j])    rst[i+1][j+1] = rst[i][j] + 1;   else    rst[i+1][j+1] = max(rst[i][j+1],rst[i+1][j]); return rst[len1][len1];}int main(){ //ifstream cin("test.txt"); int n; string s1,s2; while(cin>>n) {  cin>>s2; s1 = s2;  reverse(s2.begin(),s2.end());  cout<<s1.length() - LCS(s1,s2)<<endl; } return 0;}