POJ 1159 Palindrome

实质是：求一个字符串与其逆字符串的LCS.

LCS用short int开数组刚好可以卡着内存过！！！ 

                                                      Palindrome

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 46566		Accepted: 15910

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

Source

IOI 2000

 

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

short int dp[5555][5555];

int main()
{
    char s1[5555],s2[5555];
    s1[0]=s2[0]=':';
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
      {
          cin>>s1[i];
          s2[i]=s1[i];
      }
    reverse(s2+1,s2+1+n);
//    cout<<s2;
    memset(dp,0,sizeof(dp));

    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
    {
        if(s1[i]==s2[j])
            dp[i][j]=dp[i-1][j-1]+1;
        else
            dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
    }

    cout<<n-dp[n][n]<<endl;

    return 0;
}
 

 当然，也可以用滚存：

11613675

CKboss

1159

Accepted

792K

1110MS

G++

635B

2013-05-19 15:51:30

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

int dp[2][5555];

int main()
{
    char s1[5555],s2[5555];
    s1[0]=s2[0]=':';
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
      {
          cin>>s1[i];
          s2[i]=s1[i];
      }
    reverse(s2+1,s2+1+n);
//    cout<<s2;
    memset(dp,0,sizeof(dp));

    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
    {
        if(s1[i]==s2[j])
            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]);
    }

    cout<<n-dp[n%2][n]<<endl;

    return 0;
}