連結：

Palindrome

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2430    Accepted Submission(s): 838


Problem 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
Recommend linle

題意：

先給你字串的長度 N  再給你 N 個字元組成的序列,問最小多少個操作【新增一些字元】可以將原來的序列變成迴文串

分析：

就是逆序存一下當前序列,然後求出 LCS 【最長公共子序列】 用 N 減去所求的最長公共子序列就可以了 注意到序列很長 5000 如果用裸 dp[5000][5000] 就 MLE 了, 以前在 POJ 上做過這道題, 記憶體是 hdu  的兩倍，看過一篇部落格是關於記憶化儲存的陣列開 dp[maxn][maxn] short int 隨便就搞過去了，然後前天晚上就一直 MLE 啊Orz  完了學妹對我說 %2 開個 dp[2][5000] 就可以過了。。。 也就是前一篇部落格的滾動陣列思想：

hdu 1159 Common Subsequence 【LCS 基礎入門】

當前這一行的狀態只由上一行的狀態和這一行前面的狀態確定，所以只有記錄兩組Orz 也就是說直接把 裸的 LCS 的陣列定義改成 dp[2][maxn], 然後把裸的模板中的 i 全部 %2 就好了 最後輸出 n - dp[n%2][n]

code：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

const int maxn = 5005;
int dp[2][maxn];
char s1[maxn], s2[maxn];

void LCS(int len1,int len2)
{
    for(int i = 1; i <= len1; i++)
    {
        for(int j = 1; j <= len2; j++)
        {
            if(s1[i-1] == s2[j-1]) dp[i%2][j] = dp[(i-1)%2][j-1]+1;
            else
            {
                int m1 = dp[(i-1)%2][j];
                int m2 = dp[i%2][j-1];
                dp[i%2][j] = max(m1, m2);
            }
        }
    }
}

int main()
{
    int n;
    while(scanf("%d", &n) != EOF)
    {
        scanf("%s", s1);
        for(int i = 0; i < n; i++)
        s2[i] = s1[n-i-1];
            memset(dp,0,sizeof(dp));
        LCS(n,n);
        printf("%d\n", n-dp[n%2][n]);
    }
    return 0;
}