LightOJ 1044 Palindrome Partitioning(简单字符串DP)

A palindrome partition is the partitioning of a string such that each separate substring is a palindrome.

For example, the string "ABACABA" could be partitioned in several different ways, such as {"A","B","A","C","A","B","A"}, {"A","BACAB","A"}, {"ABA","C","ABA"}, or {"ABACABA"}, among others.

You are given a string s. Return the minimum possible number of substrings in a palindrome partition of s.

Input

Input starts with an integer T (≤ 40), denoting the number of test cases.

Each case begins with a non-empty string s of uppercase letters with length no more than 1000.

Output

For each case of input you have to print the case number and the desired result.

Sample Input

AAAA

ABCDEFGH

QWERTYTREWQWERT

Sample Output

Case 1: 1

Case 2: 8

Case 3: 5

题意：给出一个字符串，将该字符串切割成若干个回文串，使切割后的回文串数最小，求这个回文串数

题解：这道题其实在之前DP百题大过关中已经出现过了，复杂度是n^3的，今天写的时候突然糊出了n^2的做法，再记录一下

首先是状态转移方程的推导

f[i]表示1-i之间的最小分割数，很显然只有当i~j是回文串的时候i-1才能转移到j，贡献为1

所以状态转移方程为

f[j]=max{f[i-1]+1}（i~j为回文串）

这个转移方程是n^2的，但上次写的时候在里面加入了O(N)的回文串判断，复杂度变成了O(N^3)

如今想想，其实是不是回文串这东西是可以预处理的

枚举每一个中点，向两边拓展，可以在O(N^2)中处理出每一段i~j是否是回文串

然后就有了O(N^2)复杂度的DP

代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

int c[1010][1010],n,f[1010];
char s[1010];

void get(int l,int r)
{
    while(l>0&&r<=n)
    {
        if(s[l]==s[r])
        {
            c[l][r]=1;
        }
        else
        {
            return ;
        }
        l--;
        r++;
    }
}

int main()
{
    int t,ttt=0;
    scanf("%d",&t);
    while(t--)
    {
        ttt++;
        memset(f,0x3f,sizeof(f));
        memset(c,0,sizeof(c));
        scanf("%s",s+1);
        n=strlen(s+1);
        for(int i=1; i<=n; i++)
        {
            get(i,i);
            get(i,i+1);
        }
        f[0]=0;
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=i; j++)
            {
                if(c[j][i])
                {
                    f[i]=min(f[i],f[j-1]+1);
                }
            }
        }
        printf("Case %d: %d\n",ttt,f[n]);
    }

}

大佬们的常数都好优越啊，N^3跑的都比N^2快qwq

LightOJ 1044 Palindrome Partitioning(简单字符串DP)

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