UVa 11584 - Partitioning by Palindromes 回文串dp

Problem H: Partitioning by Palindromes

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 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

------------------------
我又智硬了吗orz

f[i]=min(f[j-1]+1) ( j<i,rev[j,i])

-----------------------

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int OO=1e9;
int n;
char s[1111];
int f[1111];
bool rev[1111][1111];

int main()
{
int T;
scanf("%d",&T);
while (T--)
{
memset(rev,0,sizeof(rev));
scanf("%s",s+1);
n=strlen(s+1);
for (int i=1;i<=n;i++)
{
rev[i][i]=true;
if (s[i]==s[i+1]&&i+1<=n)
{
rev[i][i+1]=true;
}
}
for (int k=2;k<=n;k++)
{
for (int i=1;i+k<=n;i++)
{
if (s[i]==s[i+k]&&rev[i+1][i+k-1])
{
rev[i][i+k]=true;
}
}
}
for (int i=1;i<=n;i++) f[i]=OO;
f[0]=0;
for (int i=1;i<=n;i++)
{
for (int j=1;j<=i;j++)
{
if (rev[j][i])
{
f[i]=min(f[i],f[j-1]+1);
}
}
}
int ans=f
;
printf("%d\n",ans);
}
return 0;
}