uva 11584 - Partitioning by Palindromes （dp）

题目链接http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2631

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

 

 

dp入门题，可归为序列问题。即

给一行序列，要求满足条件的最优解

这类问题很多都可化为子问题，从头向后走到的第i个时，前面序列的最优解

有时这个dp[i]只与dp[i-1]相关，有时与1 ~ i-1都相关

 

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>

using namespace std;
int dp[1005];
char text[1005];
int Can(int l,int r){
    int tag = 1;
    for(int i = 0;i < (r-l)/2+1;i++){
        if(text[l+i] != text[r-i]){
            tag = 0;
            break;
        }
    }
    return tag;
}

int main(){
    int n;

    cin >> n;
    while(n--){
        text[0] = '0';
        cin >> text+1;
        dp[0] = 0;
        int len = strlen(text)-1;
        for(int i = 1;i <= len;i++){
            int minx = 10000,tem;
            for(int j = 0;j < i;j++){
                if(Can(j+1,i))   tem = dp[j] + 1;
                else    tem = dp[j] + i - j;
                minx = min(minx,tem);
            }
            dp[i] = minx;
        }
        cout << dp[len] << endl;
    }
    return 0;
}