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 ",d[m]); 34 } 35 return 0; 36 }