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：
1 #include<cstdio>
2 #include<cstring>
3 using namespace std;
4 #define MAXN 1010
5 char a[MAXN];
6 int d[MAXN];
7 int min(int x,int y)
8 {
9 return x<y?x:y;
10 }
11 bool level(int l,int r)
12 {
13 int m=(l+r)/2;
14 for(int i=l; i<=m; i++)
15 if(a[i]!=a[r-i+l]) return false;
16 return true;
17 }
18 int main()
19 {
20 int n;
21 scanf("%d",&n);
22 while(n--)
23 {
24 scanf("%s",a+1);
25 int m=strlen(a+1);
26 d[0]=0;
27 for(int i=1; i<=m+1; i++)
28 d[i]=1010;
29 for(int i=1; i<=m; i++)
30 for(int j=1; j<=i; j++)
31 if(level(j,i))
32 d[i]=min(d[i],d[j-1]+1);
33 printf("%d\n",d[m]);
34 }
35 return 0;
36 }