【HDU3294】Girls' research Manacher

最新推荐文章于 2020-07-24 19:41:20 发布

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最新推荐文章于 2020-07-24 19:41:20 发布

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本文介绍了一种解决寻找字符串中最长回文子串问题的算法实现，通过Manacher算法进行高效求解，并提供了完整的代码示例。该算法通过预处理字符串并利用已有的回文性质来减少重复计算。

Girls’ research

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 3745 Accepted Submission(s): 1402

Problem Description

One day, sailormoon girls are so delighted that they intend to research about palindromic strings. Operation contains two steps:
First step: girls will write a long string (only contains lower case) on the paper. For example, “abcde”, but ‘a’ inside is not the real ‘a’, that means if we define the ‘b’ is the real ‘a’, then we can infer that ‘c’ is the real ‘b’, ‘d’ is the real ‘c’ ……, ‘a’ is the real ‘z’. According to this, string “abcde” changes to “bcdef”.
Second step: girls will find out the longest palindromic string in the given string, the length of palindromic string must be equal or more than 2.

Input

Input contains multiple cases.
Each case contains two parts, a character and a string, they are separated by one space, the character representing the real ‘a’ is and the length of the string will not exceed 200000.All input must be lowercase.
If the length of string is len, it is marked from 0 to len-1.

Output

Please execute the operation following the two steps.
If you find one, output the start position and end position of palindromic string in a line, next line output the real palindromic string, or output “No solution!”.
If there are several answers available, please choose the string which first appears.

Sample Input

b babd
a abcd

Sample Output

0 2
aza
No solution!

题解

最长回文子串的裸题，我们只需要计算出左右端点就可以，注意我们为了消除奇数偶数的影响,变成了 $T[]$ ,所以要注意下标的变化。

代码

#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
const int MAXN=1e6;
using namespace std;
int p[MAXN];
int Left,Right,Mid;
char s[MAXN],T[MAXN];
inline int Init()
{
    memset(T,0,sizeof(T));
    int len=strlen(s);
    int j=2;
    T[0]='$';
    T[1]='#';
    for(int i=0;i<len;i++)
    {
        T[++j]=s[i];
        T[++j]='#';
    }
    T[j]='\0';
    return j;
}
inline bool Manacher()
{
    memset(p,0,sizeof(p));
    int len=Init();
    int maxx=-1;
    int mx=0,id;
    for(int i=1;i<=len;i++)
    {
        if(i<mx)
            p[i]=min(mx-i,p[2*id-i]);
        else 
            p[i]=1;
        while(T[i+p[i]]==T[i-p[i]]&&i+p[i]<=len)
            p[i]++;
        if(i+p[i]>mx)
        {
            id=i;
            mx=i+p[i];
        }
        maxx=max(maxx,p[i]-1);
    }
    cout<<id<<" "<<endl;
    cout<<maxx;
    Left=(id-2)/2+1-maxx/2;
    Right=(id-2)/2+1+maxx/2; 
    if(maxx<2) 
        return 0;
    else
        return 1;
}
int main()
{
    char ch;
    int n;
    cin>>n;
    while(n--)
    {
        scanf("%s",s);
        if(Manacher()) 
            printf("%d %d\n",Left,Right);
        else
            printf("No solution\n");
    }
    return 0;
}