POJ 1159 Palindrome(DP LCS&滚动数组)

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本文探讨如何通过最少字符插入操作将任意字符串转化为回文串，并提供了两种解法：一种是利用最长公共子序列（LCS）的概念，另一种则是采用滚动数组优化空间复杂度。

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

题目大意：求插入最少的字符使得串为palindeome（回文串）。

解法一：遇到问题要剖析其原型，此题就是正序和逆序的LCS，结果来n-commenlength.

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#define LL long long
#define inf 0x3f3f3f3f
using namespace std;
short dp[5001][5001];
char a[5005],b[5005];
int main()
{
    int n,m,x,k,i,j;
    int cla;
    while(~scanf("%d",&n))
    {
        int s=0;
        dp[0][0]=0;<span id="transmark"></span>
        scanf("%s",a);
        for(i=n-1;i>=0;i--)
        b[s++]=a[i];
        b[s]='\0';
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                if(a[i-1]==b[j-1])
                {
                    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;
}

解法二：利用滚动数组来减少空间的开支。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<queue>
#include<map>
#include<stack>
#define L1 long long
#define L2 int
#define inf 0x3f3f3f3f
using namespace std;
const int m1=1001000;
const int m2=1010;
L2 dp[3][5010],a[100010];
char s[5010],s1[5010];
int main()
{
    int n,m,i,j,k;
    while(~scanf("%d",&n)){
            getchar();
        scanf("%s",s+1);int t=1;
        for(i=n;i>=1;--i){
            s1[t++]=s[i];
        }
        s1[t]='\0';
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;++i){
            for(j=1;j<=n;++j){
                if(s[i]==s1[j]){
                    dp[1][j]=dp[0][j-1]+1;
                }
                else{
                    dp[1][j]=max(dp[0][j],dp[1][j-1]);
                }
            }
            for(j=1;j<=n;j++){
                dp[0][j]=dp[1][j];
            }
        }
        printf("%d\n",n-dp[0][n]);
    }
    return 0;
}