G - dp-easy

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最新推荐文章于 2025-06-26 20:07:05 发布

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分类专栏：动态规划

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本文介绍了一道经典的动态规划问题——如何将字符串最少分割成回文子串，并提供了完整的代码实现及解析。

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G - dp-easy

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Time Limit: 1000MS Memory Limit: 0KB 64bit IO Format: %lld & %llu

Submit Status

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

这是一道动态规划基本题，还没学到DP，不过在师姐的大力讲解下，总算是明白个差不多了。

#include<stdio.h>
#include<string.h>
const int INF=(1<<29);
char z[1007];
int dp[1007];

bool judge(int left,int right)
{
      for(int c=left,d=right;c<=d;++c,--d)
      if(z[c]!=z[d])return false;
      return true;

}

int min(int x,int y)
{
   return x<y?x:y;
}

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        while(n--)
        {
            scanf("%s",z+1);
            dp[0]=0;
            int len=strlen(z+1);
            for(int i=1;i<=len;++i)
            dp[i]=INF;
            for(int a=1;a<=len;++a)
            for(int b=1;b<=a;++b)
            if(judge(b,a))
            dp[a]=min(dp[a],dp[b-1]+1);
            printf("%d\n",dp[len]);
        }
    }
    return 0;
}