Poj1159--palindrome (DP: Longest common Subsequence variant + scrolling array)

Palindrome

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 53414		Accepted: 18449

Description

A palindrome is a symmetrical string, which is, a string read identically from the left to the right as well as from the right to left. You-to-write a program which, given a string, determines the minimal number of characters to being inserted into the STR ing in order to obtain a palindrome. 

As an example, by inserting 2 characters, the string "ab3bd" can is transformed into a palindrome ("Dab3bad" or "Adb3bda") . However, inserting fewer than 2 characters does not produce a palindrome. 

Input

Your program was 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 is to be considered distinct.

Output

Your program is-to-write to standard output. The first line contains one integer and which is the desired minimal number.

Sample Input

5ab3bd

Sample Output

2

Source

IOI 2000 topic: Adding the fewest characters to make a palindrome string if a string is a palindrome, then it and its inverse array of the longest common sub-sequence of its own length, so that the longest common sub-sequence of the length of the total length minus it, the result is to modify the length.

Array of 5000*5000 using short

#include <cstdio> #include <cstring> #include <algorithm>using namespace Std;short dp[5100][5100]; Char str1[5100], str2[5100]; int main () {    int i, j, N;    while (scanf ("%d", &n)!=eof)    {        scanf ("%s", str1);        for (i = 0; i < n; i++)            str2[n-1-i] = str1[i];        Str2[i] = ' + ';        for (i = 1; I <= N, i++) for            (j = 1; J <= N; j + +)            {                if (str1[i-1] = str2[j-1])                    dp[i][j] = dp[i -1][j-1]+1;                else                    dp[i][j] = max (dp[i-1][j],dp[i][j-1]);            }        printf ("%d\n", N-dp[n][n]);    }    return 0;}

Using a scrolling array

Scrolling arrays: Using two rows of arrays to simulate large two-dimensional arrays

#include <cstdio> #include <cstring> #include <algorithm>using namespace Std;int dp[2][5100]; char STR1[5100], str2[5100]; int main () {    int i, J, N, K;    while (scanf ("%d", &n)!=eof)    {        scanf ("%s", str1);        for (i = 0; i < n; i++)            str2[n-1-i] = str1[i];        Str2[i] = ' + ';        k = 0;        for (i = 1; I <= n; i++)        {            k = 1-k;            for (j = 1; J <= N; j + +)            {                if (str1[i-1] = = str2[j-1])                    dp[k][j] = dp[1-k][j-1]+1;                else                    dp[k][j] = max (dp[1-k][j],dp[k][j-1]);            }        }        printf ("%d\n", N-dp[k][n]);    }    return 0;}

Poj1159--palindrome (DP: Longest common Subsequence variant + scrolling array)