动态规划——H 最少回文串

We say a sequence of characters is a palindrome if it is the same written forwards and backwards. For example, ‘racecar’ is a palindrome, but ‘fastcar’ 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

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.

Output

For each test case, output a line containing the minimum number of groups required to partition the input into groups of palindromes.

Sample Input

3

racecar

fastcar

aaadbccb

Sample Output

1

7

3

题目大意：一串字符串中，找出最少组成字符串

解题思路：

用枚举法枚举起点到终点是否是回文串，即判断j-i是否是回文串，如果是单个的字母，则也单独组成一个回文串

程序代码：

 #include<cstdio>
 #include<cstring>
 using namespace std;
 #define MAXN 1010
 char a[MAXN];
 int d[MAXN];
 int min(int x,int y)
 {
     return x<y?x:y;
 }
 bool level(int l,int r)
 {
     int m=(l+r)/;
     for(int i=l; i<=m; i++)
         if(a[i]!=a[r-i+l]) return false;
     return true;
 }
 int main()
 {
     int n;
     scanf("%d",&n);
     while(n--)
     {
         scanf("%s",a+);
         int m=strlen(a+);
         d[]=;
         for(int i=; i<=m+; i++)
             d[i]=;
         for(int i=; i<=m; i++)
             for(int j=; j<=i; j++)
                 if(level(j,i))
                     d[i]=min(d[i],d[j-]+);
         printf("%d\n",d[m]);
     }
     return ;
 }