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

3racecarfastcaraaadbccb

Sample Output

173

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

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[n];        printf("%d\n",ans);    }    return 0;}