NYOJ练习题 Splits the string （简单动态规划）

Splits the string

时间限制： 1000 ms  |  内存限制： 65535 KB

描述

Hrdv is interested in a string,especially the palindrome string.So he wants some palindrome string.

A sequence of characters is a palindrome if it is the same written forwards and backwards. For example, 'abeba' is a palindrome, but 'abcd' 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.

输入

Each test case consists of a single line of between 1 and 1000 lowercase letters, with no whitespace within.

输出

For each test case, output a line containing the minimum number of groups required to partition the input into groups of palindromes.

样例输入

racecar
fastcar
aaadbccb

样例输出

1
7
3

题意：给定一个字符串，求最少需要几个回文子串组成此字符串。

解题思路：用dp[i] 记录从0到当前i位置这一段最少由几个回文子串组成，动态转移方程：dp[i] = min(dp[i],dp[j-1]+1);

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char str[1010];
int dp[1010];
bool judge(int x,int y) //判断是不是回文串
{
	while(x <= y)
	{
		if(str[x] != str[y])
			return false;
		x++;
		y--;
	}
	return true;
}
int main()
{
	int len, i, j;
	while(gets(str) != NULL)
	{
		len = strlen(str);
		for(i = 0; i < len; i++)
		{
			dp[i] = i + 1; //假设前面的都不能组成回文串
			for(j = 0; j <= i; j++)
				if(str[j] == str[i] && judge(j,i))
					dp[i] = min(dp[i], dp[j-1]+1);
		}
		printf("%d\n",dp[len-1]);
	}
	return 0;
}