D

  

                                     Palindrome Partitioning

                                                                                            Time Limit: 1000ms

                                                                                          Memory Limit: 32768KB

 

A palindrome partition is the partitioning of a string such that each separate substring is a palindrome.

For example, the string "ABACABA" could be partitioned in several different ways, such as {"A","B","A","C","A","B","A"}, {"A","BACAB","A"}, {"ABA","C","ABA"}, or {"ABACABA"}, among others.

You are given a string s. Return the minimum possible number of substrings in a palindrome partition of s.

Input

Input starts with an integer T (≤ 40), denoting the number of test cases.

Each case begins with a non-empty string s of uppercase letters with length no more than 1000.

 

Output

For each case of input you have to print the case number and the desired result.

 

Sample Input

Sample Input	Output for Sample Input
3 AAAA ABCDEFGH QWERTYTREWQWERT	Case 1: 1 Case 2: 8 Case 3: 5

 

此提找到一种代码精简的方法！！不容易啊，虽然1==Ms但是代码短了很多，也易于理解。。。。。。

#include<stdio.h>
#include<string.h>
#define N 1010
#include <algorithm>
using namespace std;
char a[N];
int f[N];
int pali (int n,int m)
{
    int i,j;
    for(i=n,j=m;i<=(n+m)/2;i++,j--)
    if(a[i]!=a[j])return 0;
    return 1;
}
int main ()
{
    int T,i,j,a_len,ca;
        scanf("%d",&T);
        for(ca=1;ca<=T;ca++)
        {
            scanf("%s",&a);
            a_len=strlen(a);
            for(i=0;i<a_len;i++)
            { f[i]=i+1;
              for(j=0;j<=i;j++)
              if(pali(j,i))
              f[i]=min(f[i],f[j-1]+1);
            } 
            printf("Case %d: ",ca);
            printf("%d
",f[a_len-1]);
        }
}