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
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
#include <queue>
#include <iostream>
#include<cstring>
#include<stdio.h>
#define maxn 1004
using namespace std;
int n,dp[maxn];
bool table[maxn][maxn];
char seq[maxn*2];int trans(int i,int j)
{if(table[i][j]) return 1;return j-i+1;
}
/*
题目大义：
给定一个字符串，问最少能分解成多少个回文串动态规划部分的思想很简单，
两重循环即可，如果预处理知道任意区间的
字符串是否为回文的即可。重点是如何预处理得知任意区间的字符串是否为回文的，
这里采用枚举中心的方法，
但普通的方法仅对奇数长度的字符串有效，
而对于偶数长度的回文字符串却不奏效。所以引入一种处理方法，
即在每两个字符之间插入一个特殊字符，
对新字符串进行扫描判断判断，
最终得出的回文区间i,j映射3到真实区间i>>1,(j-1)>>1;
(最终区间范围是0~l-1)
*/int main()
{ios::sync_with_stdio(false);int t;scanf("%d",&t);while(t--){seq[0]='#';scanf("%s",seq);int l=strlen(seq);for(int i=l;i>=0;i--) seq[(i<<1)+1]=seq[i];for(int i=0;i<=2*l-1;i+=2) seq[i]='#';memset(table,0,sizeof(table));for(int i=0;i<=2*l-1;i++){int ll=i,rr=i;while(ll>=0&&rr<=2*l-1&&seq[ll]==seq[rr]){table[ll>>1][(rr-1)>>1]=true;ll--,rr++;}}/*for(int i=0;i<l;i++)for(int j=0;j<=i;j++)if(table[j][i]) cout<<j<<" "<<i<<endl;*/dp[0]=1;for(int i=0;i<l;i++){dp[i]=trans(0,i);for(int j=0;j<i;j++){if(!table[j+1][i]) continue;dp[i]=min(dp[i],dp[j]+1);}}printf("%d\n",dp[l-1]);}return 0;
}