poj 1159 Palindrome(最长公共子序列)

/*

Palindrome

http://poj.org/problem?id=1159

Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 53898 Accepted: 18618
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
*/
/*
给你一串字符串，让你求最少加入几个字符，才能使得这个字符串是个回文串。

*/

/*

最长公共子序列的题目的应用

m=strlen(s1) n=strlen(s2);

for(i=0;i<=m;i++)

   dp[i][0]=0;

for(i=0;i<=n;i++)

   dp[0][i]=0;

for(i=1;i<=m;i++)

  for(j=1;j<=n;j++) 

     if(s1[i-1]==s2[j-1])

       dp[i][j]=dp[i-1][j-1]+1;

     else

       dp[i][j]=max(dp[i-1][j],dp[i][j-1]);

*/

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
int dp[10][5001];
int s1[5001];
int s2[5001];
int main()
{
  int n;
  while(~scanf("%d",&n))
  {
  int i,j;
  char ch;
  getchar();
  memset(dp,0,sizeof(dp));
  memset(s1,0,sizeof(s1));
  memset(s2,0,sizeof(s2));
  for(i=1;i<=n;i++)
  {
   scanf("%c",&ch);
   s1[i]=ch;
   s2[n+1-i]=ch;
}
  dp[0][1]=dp[0][0]=0; 
  for(i=0;i<=n;i++)
  dp[0][i]=0;

for(i=1;i<=n;i++)
 for(j=1;j<=n;j++)
 {
if(s1[i]==s2[j])
   dp[i%2][j]=dp[(i-1)%2][j-1]+1;//此处用dp[i][j]总是超内存 
   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;
}