UVA 题目11584 - Partitioning by Palindromes(DP)

最新推荐文章于 2018-09-02 18:11:00 发布

Jogging_Clown

最新推荐文章于 2018-09-02 18:11:00 发布

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本博客探讨如何将给定的字符序列拆分为最少数量的回文子串。

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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

题目大意：一个串能拆成几个回文串

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f3f3f3f
char str[1010];
int dp[1010],ok[1010][1010],len;
void fun()
{
    int i,j;
    for(i=1;i<=len;i++)
        for(j=1;j<=i;j++)
            if(i==j||(str[i]==str[j]&&(ok[j+1][i-1]||i-j==1)))
                ok[j][i]=1;
            else
                ok[j][i]=0;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",str+1);
        len=strlen(str+1);
        fun();
        memset(dp,INF,sizeof(dp));
        dp[0]=0;
        int i,j;
        for(i=1;i<=len;i++)
            for(j=1;j<=i;j++)
                if(ok[j][i])
                    dp[i]=min(dp[i],dp[j-1]+1);
        printf("%d\n",dp[len]);
    }
}