UVa 11584 Partitioning by Palindromes(DP 最少对称串)

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题意  判断一个串最少可以分解为多少个对称串   一个串从左往后和从右往左是一样的  这个串就称为对沉串

令d[i]表示给定串的前i个字母至少可以分解为多少个对称串  那么对于j=1~i   若(i,j)是一个对称串  那么有  d[i]=min(d[i],d[j-1]+1)   然后就得到答案了

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 1005;

int d[N]; char s[N];
bool isPal (int a, int b)
{
    while (a <= b && s[a] == s[b])
        ++a, --b;
    return a >= b;
}

int main()
{
    int cas, l;
    scanf ("%d", &cas);
    for (int ca = 1; ca <= cas; ++ca)
    {
        scanf ("%s", s + 1);
        l = strlen (s + 1);
        memset (d, 0x3f, sizeof (d));
        d[0] = 0;

        for (int i = 1; i <= l; ++i)
        {
            for (int j = 1; j <= i; ++j)
                if (isPal (j, i))  d[i] = min (d[i], d[j - 1] + 1);
        }
        printf ("%d\n", d[l]);
    }
    return 0;
}

Partitioning by Palindromes

We say a sequence of characters is a palindromeif 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 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.

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

UVa 11584 Partitioning by Palindromes(DP 最少对称串)

标签：uva   小白   acm   dp   字符串   

原文地址：http://blog.csdn.net/acvay/article/details/38820795