16-最少回文数组

　　　　　　　　　　　　　　　　　　　　　

　　　　　　　　　　　　　　　　Splits the string

　　　　　　　　　　　　　　　　　　　　时间限制：1000 ms  |  内存限制：65535 KB

　　　　　　　　　　　　　　　　　　　　　　　　　　难度：3

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

上传者

TC_胡仁东

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int dp[1005];
char a[1005];

int judge(int i, int j){

//    while(a[i] == a[j]){
//        if(i == j)
//            break;
//        i++;
//        if(i == j)
//            break;
//        j--;
//        if(i == j)
//            break;
//    }
//    if(i == j && a[i] == a[j])
//        return 1;
//    else
//        return 0;
    //上面判断回文太麻烦了
    int mid = (i + j) >> 1;
    int m = i;
    for(int i; i <= mid; i++){
        if(a[i] != a[j - i + m])  //注意j - i时要加上数组片段传来的起点. 
            return 0;
    } 
    return 1;
}

int main(){
    while(~scanf("%s", a + 1)){
        int len = strlen(a + 1);
        for(int i = 1; i <= len; i++){
            dp[i] = i;
            for(int j = 1; j <= i; j++){
                if(a[i] == a[j] && judge(j, i)){        //dp[i]表示的是0到i的最少回文串数 
                    dp[i] = min(dp[i], dp[j - 1] + 1);  //j从是从0开始扫描到i如果发现j到i是回文，
                                                        //则j到i为一个回文串，就可以得到这个递推关系 
                }
            }
        }    
        printf("%d
", dp[len]);     
    }
    return 0;
}