最长公共子序列运用

给定一个字符串s，你可以从中删除一些字符，使得剩下的串是一个回文串。如何删除才能使得回文串最长呢？
输出需要删除的字符个数。
思路：将原字符串反转，求出元字符串与反转字符串最长公共子序列(也就是最长的回文串);
c++代码(网上搜的)：

#include <bits/stdc++.h>
using namespace std;

const int MAXN=1010;
int dp[MAXN][MAXN];

class Solution
{
public:
   int solve(string &s)
   {
       return s.length()-getLCS(s);
   }

   int getLCS(string &s1)
   {
       string s2(s1);
       reverse(s2.begin(),s2.end());
       int len=s1.length();
       memset(dp,0,sizeof dp);
       for(int i=0;i<len;++i)
       {
           for(int j=0;j<len;++j)
           {
               if(s1[i]==s2[j])
                   dp[i+1][j+1]=dp[i][j]+1;
               else dp[i+1][j+1]=max(dp[i][j+1],dp[i+1][j]);
           }
       }
       return dp[len][len];
   }
};

int main()
{
   string s;
   while(cin>>s)
   {
       Solution solution;
       cout<<solution.solve(s)<<endl;
   }
   return 0;
}

c语言代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
void reverse(char *str,int from,int to)
{
    while (from < to)
    {
        char temp = str[from];
        str[from++] = str[to];
        str[to--] = temp;
    }
}
int getLcs(char *str1,char *str2,int length)
{
    //printf("%s %s\n",str1,str2);//居然是一样的
    int dp[length][length];
    int i =0,j = 0;
    for (i = 0;i <length;i++)
    {
        for (j = 0;j < length;j++)
            dp[i][j] = 0;
    }
//    int i = 0,j = 0;
    for (i = 0;i < length;i++)
    {
        for (j = 0;j < length;j++)
        {
            if (i == 0 || j == 0)
            {
                if (str1[i] == str2[j])
                {
                    dp[i][j] = 1;
                }
                else
                {
                    if (i > 0)
                        dp[i][j] = dp[i-1][j];
                    if (j > 0)
                        dp[i][j] = dp[i][j-1];
                }
            }
            else if (str1[i] == str2[j])
            {
                dp[i][j] = dp[i-1][j-1]+1;
            }
            else {
                dp[i][j] = ((dp[i-1][j] > dp[i][j-1]) ?dp[i-1][j]:dp[i][j-1]);
            }
        }
    }
    return dp[length-1][length-1];
}
int main()
{
    char str[1024];//注意下字符串指针
    //char *str1 = "abcda";
    //char *str2 = "adcba";

    while (gets(str))
    {
        int len = strlen(str);
        char str1[1024];
    //    char *str1 = str;
    //    printf("str1 = %s\n",str1);

        strcpy(str1,str);
        reverse(str,0,len-1);


        //printf("str2 = %s\n",str2);
        int result = getLcs(str1,str,len);

        printf("%d\n",(len-result));
        //free(str2);
    }
    return 0;
    //getLcs()

}