UVA-11584

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

题目大意：问最少的回文串的总数。

思路：开始我想用manacher算法。只需要到时候处理一下就可以。但是还是想简单了。折腾了两个小时都没搞好。果断弃坑。ps：按理说应该是可以的，可能是我太马虎了。

赛后发现原来是dp，真是藏的好深啊。。。

因为题目数据是1000。时间限制3s。我们就用比较好理解的O(n3)的复杂度的方法吧。

dp[ i ]表示前i个字符能构成的最小回文串总数。

代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define INF 999999
using namespace std;
char a[1005];
bool IsHuiWen(int l,int r){
    for(int j=l,i=r;j<=i;j++,i--){
        if(a[j]!=a[i])
            return false;
    }
    return true;
}
int main(){
    int T;
    cin>>T;
    while(T--){
        int dp[1005];
        scanf("%s",a+1);
        int len=strlen(a+1);
        for(int i=1;i<=len;i++){
            dp[i]=INF;
        }
        dp[0]=0;
        for(int i=1;i<=len;i++){
            for(int j=1;j<=i;j++){
                if(IsHuiWen(j,i)){
                    dp[i]=min(dp[i],dp[j-1]+1); //如果j到i是回文串，那么我们就可以利用dp[j-1]的最少回文串来知道dp[i]的最少回文串。
                }
            }
        }
        cout<<dp[len]<<endl;
    }
    return 0;
}