POJ 1159 Palindrome（lcs加滾動數組）

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Palindrome

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 52350		Accepted: 18041

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

設原序列S的逆序列為S‘ ，則這道題目的關鍵在於，

最少需要補充的字母數 = 原序列S的長度 —  S和S‘的最長公用子串長度

剛開始做的時候沒有想到滾動數組，結果就mle了。。（後來聽說改成short int ，結果就ac了）空間開銷很大

#include <iostream>#include <cstring>using namespace std;#define maxn 5005char str[maxn];//輸入的字串char nstr[maxn];//輸入的字串的逆序列short int dp[maxn][maxn];//定義動態二維數組並初始化int main(){    int n;//輸入的字串長度    cin>>n;    dp[0][0]=0;    memset(dp,0,sizeof(dp));    int i,j;    for(i=1,j=n;i<=n&&j>=1;i++,j--)    {        cin>>str[i];        nstr[j]=str[i];//給逆序列賦值    }    for(i=1;i<=n;i++)        for(j=1;j<=n;j++)        {            if(str[i]==nstr[j])//如果字元相等，dp[i][j]等於左上值加1                dp[i][j]=dp[i-1][j-1]+1;            if(str[i]!=nstr[j])//不相等，dp[i][j]等於上方和左方dp值得最大值                dp[i][j]=dp[i-1][j]>dp[i][j-1]?dp[i-1][j]:dp[i][j-1];        }cout<<n-dp[n][n]<<endl;return 0;}

滾動數組

滾動數組的作用在於最佳化空間，主要應用在遞推或動態規劃中（如01背包問題）。因為DP題目是一個自底向上的擴充過程，我們常常需要用到的是連續的解，前面的解往往可以捨去。所以用滾動數組最佳化是很有效。利用滾動數組的話在N很大的情況下可以達到壓縮儲存的作用。

例：斐波那契數列：

一般代碼

int fib(int n)  {      Fib[0] = 0;      Fib[1] = 1;      Fib[2] = 1;      for(int i = 3; i <= n; ++i)          Fib[i] = Fib[i - 1] + Fib[i - 2];      return Fib[n];  } 

應用滾動數組

int fib(int n)  {      Fib[1] = 0;       Fib[2] = 1;      for(int i = 2; i <= n; ++i)      {          Fib[0] = Fib[1];           Fib[1] = Fib[2];          Fib[2] = Fib[0] + Fib[1];      }      return Fib[2];  } 

對於本題,用行數以0  1  0  1的滾動方式，滾動運算式為i%2和(i-1)%2 ，求餘滾動

#include <iostream>#include <cstring>using namespace std;#define maxn 5005char str[maxn];//輸入的字串char nstr[maxn];//輸入的字串的逆序列int dp[2][maxn];//定義動態二維數組並初始化int main(){    int n;//輸入的字串長度    cin>>n;    dp[0][0]=0;    memset(dp,0,sizeof(dp));    int i,j;    for(i=1,j=n;i<=n&&j>=1;i++,j--)    {        cin>>str[i];        nstr[j]=str[i];//給逆序列賦值    }    for(i=1;i<=n;i++)        for(j=1;j<=n;j++)        {            if(str[i]==nstr[j])//如果字元相等，dp[i][j]等於左上值加1                dp[i%2][j]=dp[(i-1)%2][j-1]+1;            if(str[i]!=nstr[j])//不相等，dp[i][j]等於上方和左方dp值得最大值                dp[i%2][j]=dp[(i-1)%2][j]>dp[i%2][j-1]?dp[(i-1)%2][j]:dp[i%2][j-1];        }cout<<n-dp[n%2][n]<<endl;return 0;}