Palindrome Time Limit:3000MS Memory Limit:65536K
Total Submit:12142 Accepted:4213
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
Source
IOI 2000
分析：
动态规划求解。
设ch[1]..ch[n]表示字符串1至n位，i为左游标，j为右游标 ，则i从n递减，j从i开始递增。
min[i][j]表示i和j之间至少需要插入多少个字符才能对称,我们最终需要得到的值是min[1][n].
则
if(ch[i]==ch[j])
min[i][j]=min[i+1][j-1];
else
min[i][j] = 1 + （min[i+1][j]和min[i][j-1]中的较小值）;
另外，min[][]可以定义为short 而非 int,动态规划算法通常紧缺memory 。
什么？你说short 和 int是一样的？ 呵呵，兄弟，大概你看的是年代比较久远的C语言教材吧～
1
#include "string.h"
2#include "stdio.h"
3 int min[5001][5001];
4
5int MIN(short a,short b)
6
{
7
if(a>b)return b;
8 else return a;
9
}
10
11int main()
12{
13 int n;
14 int i,j;
15 char ch[5001];
16
17
18 scanf("%d",&n);
19 scanf("%s",ch+1);
20 for(i=1;i<=n;i++)
21 for(j=1;j<=n;j++)
22 min[i][j]=0;
23
24 for(i=n-1;i>=0;i--)
25 for(j=i;j<=n;j++)
26 if(ch[i]==ch[j])min[i][j]=min[i+1][j-1];
27 else min[i][j]= 1 + MIN(min[i+1][j],min[i][j-1]);
28 printf("%d",min[1][n]);
29
30return 0;
31
32}