UVA 11584 Partitioning by Palindromes(메모 DP, 레벨 4)

A - Partitioning by Palindromes
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Time Limit:1000MS    Memory Limit:0KB     64bit IO Format:%lld & %llu
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Appoint description: System Crawler (2013-05-30)
Description

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

Kevin Waugh
 
한 꿰미는 적어도 몇 개의 회문 꿰미로 나누어야 한다
사고방식: 먼저 회문열을 표시한 다음DP를 한 번 훑어본다.
   

#include<cstring>
#include<cstdio>
#include<iostream>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
using namespace std;
const int mm=1009;
char s[mm];
bool isp[mm][mm];
int dp[mm];
int main()
{
  int cas;
  while(~scanf("%d",&cas))
  {
    while(cas--)
    { clr(isp,0);
      scanf("%s",s);
      int len=strlen(s);
      FOR(i,0,len)isp[i+1][i]=isp[i+2][i]=1;
      for(int i=len;i>0;--i)
        FOR(j,i,len)
        if(isp[i+1][j-1]&&s[i-1]==s[j-1])isp[i][j]=true;
      dp[0]=0;
      FOR(i,1,len)
      {
        dp[i]=mm;
        FOR(j,1,i)
        if(isp[j][i])
          dp[i]=min(dp[i],dp[j-1]+1);
      }
      printf("%d
",dp[len]);
    }
  }
  return 0;
}