Uva--11584（动规，枚举）

2014-08-27 11:43:18

Problem H: Partitioning by Palindromes

We say a sequence of characters is apalindrome 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

 

思路：dp[i]表示把前 i 个字符划分成最少回文串的个数。dp[i] = min(dp[j] + 1) （0 <= j < i，s[j+1 ~ i]是回文串），枚举型递推动规

/*************************************************************************
    > File Name: 11584.cpp
    > Author: Nature
    > Mail: 564374850@qq.com 
    > Created Time: Tue 26 Aug 2014 09:23:34 PM CST
************************************************************************/

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;
const int INF = 1 << 29;

int n;
int len;
int dp[1005];
int judge[1005][1005];
char s[1005];

void Pre_process(){
    memset(judge,0,sizeof(judge));
    for(int i = 1; i <= len; ++i){
        for(int j = 0; j < len; ++j){
            if(i - j < 1 || i + j > len || s[i - j] != s[i + j])
                break;
            judge[i - j][i + j] = 1;
        }
        for(int j = 0; j < len; ++j){
            if(i - j - 1 < 1 || i + j > len || s[i - j - 1] != s[i + j])
                break;
            judge[i - j - 1][i + j] = 1;
        }
    }
}

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