Splits the string

Splits the string
时间限制：1000 ms | 内存限制：65535 KB
难度：3
描述
Hrdv is interested in a string,especially the palindrome string.So he wants some palindrome string.

A sequence of characters is a palindrome if it is the same written forwards and backwards. For example, ‘abeba’ is a palindrome, but ‘abcd’ is not.

A partition of a sequence of characters is a list of one or more disjoint non-empty groups of consecutive characters whose concatenation yields the initial sequence. For example, (‘race’, ‘car’) is a partition of ‘racecar’ into two groups.

Given a sequence of characters, we can always create a partition of these characters such that each group in the partition is a palindrome! Given this observation it is natural to ask: what is the minimum number of groups needed for a given string such that every group is a palindrome?

For example:

‘racecar’ is already a palindrome, therefore it can be partitioned into one group.
‘fastcar’ does not contain any non-trivial palindromes, so it must be partitioned as (‘f’, ‘a’, ‘s’, ‘t’, ‘c’, ‘a’, ‘r’).
‘aaadbccb’ can be partitioned as (‘aaa’, ‘d’, ‘bccb’).
Input begins with the number n of test cases. Each test case consists of a single line of between 1 and 1000 lowercase letters, with no whitespace within.

输入
Each test case consists of a single line of between 1 and 1000 lowercase letters, with no whitespace within.
输出
For each test case, output a line containing the minimum number of groups required to partition the input into groups of palindromes.
样例输入
racecar
fastcar
aaadbccb
样例输出
1
7
3

我的代码：

#include <iostream>#include <cstring>using namespace std;#define min(a,b) a>b?b:a char str[1001];int dp[1001];int judge(int x,int y)//判断是否是回文串{    while (x<=y)    {        if(str[x]!=str[y])return 0;//不是就返回0        x++;y--;    }    return 1;}int main(){    int i,j,k;    while (cin >> str)    {        k=strlen(str);        i=0;        while (i<k)        {            dp[i]=i+1;//最坏的情况为字符自身长度            for (j=0;j<=i;++j)            {                if(str[j]==str[i]&&judge(j,i))                    dp[i]=min(dp[i],dp[j-1]+1);//后一个的状态需要前一个决定            }            i++;        }        cout << dp[k-1] << endl;    }    return 0;}