POJ1159 Palindrome #最长公共子序列 滚动数组#

Palindrome Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 71974   Accepted: 24995 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

卡内存，可以开 short 数组以节省空间。

// Version 1
#include 
#include 
using namespace std;

short getLcs(const string& a, const string& b)
{
    size_t n = a.length(), m = b.length();
    short** dp = new short*[n + 1];
    for (int i = 0; i <= n; i++)
    {
        dp[i] = new short[m + 1];
        for (int j = 0; j <= m; j++) dp[i][j] = 0;
    }
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            dp[i][j] = a[i - 1] == b[j - 1] ? dp[i - 1][j - 1] + 1 : max(dp[i - 1][j], dp[i][j - 1]);
    return dp[n][m];
}

int main()
{
    int n;
    cin >> n;
    string a;
    cin >> a;
    string b(a.rbegin(), a.rend());
    cout << a.length() - getLcs(a, b) << endl;
    return 0;
}

可以使用滚动数组节省空间。

// Version 2
#include 
#include 
using namespace std;

int getLcs(const string& a, const string& b)
{
    size_t n = a.length(), m = b.length();
    int** dp = new int*[2];
    for (int i = 0; i < 2; i++)
    {
        dp[i] = new int[m + 1];
        for (int j = 0; j <= m; j++) dp[i][j] = 0;
    }
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            dp[i & 1][j] = a[i - 1] == b[j - 1] ? dp[(i + 1) & 1][j - 1] + 1 : max(dp[(i + 1) & 1][j], dp[i & 1][j - 1]);
    return dp[n & 1][m];
}

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("input.txt", "r", stdin);
#endif
    int n;
    cin >> n;
    string a;
    cin >> a;
    string b(a.rbegin(), a.rend());
    cout << a.length() - getLcs(a, b) << endl;
    return 0;
}