UVA 题目11584 - Partitioning by Palindromes(DP)

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 0x3f3f3f3f3f3f3fchar 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]);    }}