问题：1159Palindrome

http://acm.pku.edu.cn/JudgeOnline/problem?id=1159

tle的代码：
   #include  < iostream >
#include  < stdio.h >
using   namespace  std;
short  a[ 5010 ][ 5010 ]  =   { 0 } ;
char  p[ 5010 ];
int  main()
{
     int  l;
     int  i, j , t, k  =   1 ;
     while (scanf( " %d%s " , & l,p) != EOF)
     {
        k  =   1 ;
         // 下面for语句产生l=5000的测试数据；
  /**/ /*        l = 5000;
        for(i = 0; i < 5000; i++)
            p[i] = 'b';
        p[0] = '1';
        p[i] = '\0';
         */
         for (t  =   1 ; t  <  l; t ++ )
         {
            for (i  =   0 ;i < l - t;i ++ )
            {
               j  =  i  +  t;
                if (p[i]  ==  p[j]) {
                     // k++;
                    // 把下面的换成k++对l=5000这组数据时间就不用那么多了，为什么啊？不都是一次运算么,运算时间不是相同的？
                   a[i][j]  =  a[i + 1 ][j - 1 ];
               }
                else {
                     if (a[i + 1 ][j] < a[i][j - 1 ])a[i][j] = a[i + 1 ][j] + 1 ;
                     else  a[i][j] = a[i][j - 1 ] + 1 ;
               }

           } ;
        } ;
        printf( " %d\n " , a[ 0 ][l - 1 ]);
    }
     return   0 ;
}

ac的代码：

#include  < stdio.h >
int  main()
{
    int a[2][5002]={0};
    char p[5002];
    char ch[5002];
    int i,j,n,t;
    int current;
    int before;
    scanf("%d%s",&n,p+1);

        current=1;
        before = 0;
        for(i=1;i<=n;i++)
           ch[i]=p[n-i+1];
           ch[i]='\0';
     //   printf("%s %s\n",p+1,ch+1);
        for(i=1;i<=n;i++)
        {    for(j=1;j<=n;j++)
                if(ch[i]==p[j])
                   a[current][j]=a[before][j-1]+1;
                else
                    if(a[current][j-1]>=a[before][j])a[current][j]=a[current][j-1];
                    else a[current][j]=a[before][j];
              t=current;
              current=before;
              before=t;
        }
        printf("%d\n",n-a[t][n]);
    return 0;
}