动态规划——H 最少回文串

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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 }

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动态规划——H 最少回文串

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原文地址：http://www.cnblogs.com/www-cnxcy-com/p/4722177.html