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)/2;
    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+1);
        int m=strlen(a+1);
        d[0]=0;
        for(int i=1; i<=m+1; i++)
            d[i]=1010;
        for(int i=1; i<=m; i++)
            for(int j=1; j<=i; j++)
                if(level(j,i))
                    d[i]=min(d[i],d[j-1]+1);
        printf("%d
",d[m]);
    }
    return 0;
}