专题测试三 贪心与动态规划 E - Partitioning by Palindromes

1. 题目
   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
2. 思路
   dp，dp[i]存储从第一个到第i个字母中回文数量，每次检查（j,i）一段，如果这一段是回文串，那么就有dp[i]=min(dp[i],dp[j-1]+1)
3. 代码
   

   #include<cstdio>
   #include<iostream>
   #include<string>
   #include<cstring>
   #include<algorithm>
   using namespace std;
   
   char str[1010];
   int f[1010];
   
   bool judge(int x,int y)
   {
   	while(x <= y)
   	{
   		if(str[x] != str[y])
   			return 0;
   		x++;
   		y--;
   	}
   	return 1;
   }
   
   int main()
   {
   	int len, i, j, n;
   	scanf("%d",&n);
   	while(n--)
   	{
   		scanf("%s",str);
   		len = strlen(str);
   		f[0]=1;
   		for(i = 0; i < len; i++)
   		{
   			f[i] = 1145141919;
   			for(j = 0; j <= i; j++)
   				if(judge(j,i))
   					f[i] = min(f[i], f[j-1]+1);
   		}
   		printf("%d\n",f[len-1]);
   	}
   	return 0;
   }