URAL-1297 Palindrome (最长回文子串)

Palindrome

Description

The “U.S. Robots” HQ has just received a rather alarming anonymous letter. It states that the agent from the competing «Robots Unlimited» has infiltrated into “U.S. Robotics”. «U.S. Robots» security service would have already started an undercover operation to establish the agent’s identity, but, fortunately, the letter describes communication channel the agent uses. He will publish articles containing stolen data to the “Solaris” almanac. Obviously, he will obfuscate the data, so “Robots Unlimited” will have to use a special descrambler (“Robots Unlimited” part number NPRx8086, specifications are kept secret).

Having read the letter, the “U.S. Robots” president recalled having hired the “Robots Unlimited” ex-employee John Pupkin. President knows he can trust John, because John is still angry at being mistreated by “Robots Unlimited”. Unfortunately, he was fired just before his team has finished work on the NPRx8086 design.

So, the president has assigned the task of agent’s message interception to John. At first, John felt rather embarrassed, because revealing the hidden message isn’t any easier than finding a needle in a haystack. However, after he struggled the problem for a while, he remembered that the design of NPRx8086 was still incomplete. “Robots Unlimited” fired John when he was working on a specific module, the text direction detector. Nobody else could finish that module, so the descrambler will choose the text scanning direction at random. To ensure the correct descrambling of the message by NPRx8086, agent must encode the information in such a way that the resulting secret message reads the same both forwards and backwards. 
In addition, it is reasonable to assume that the agent will be sending a very long message, so John has simply to find the longest message satisfying the mentioned property.

Your task is to help John Pupkin by writing a program to find the secret message in the text of a given article. As NPRx8086 ignores white spaces and punctuation marks, John will remove them from the text before feeding it into the program.

Input

The input consists of a single line, which contains a string of Latin alphabet letters (no other characters will appear in the string). String length will not exceed 1000 characters.

Output

The longest substring with mentioned property. If there are several such strings you should output the first of them.

Sample Input

ThesampletextthatcouldbereadedthesameinbothordersArozaupalanalapuazorA

ArozaupalanalapuazorA

/*
 * URAL 1297 Palindrome
 * 求一个字符串的最长回文子串
 *
 * 考虑用后缀数组做，首先在字符串后面加一个不会出现的字符，再将原串反转接到原串的后面，
 * 得到一个长度为2*len+1的新字符串，求出后缀数组和height数组
 * 考虑这个性质，如果一个字符串有回文子串，那么按照上面构造的新字符串的对称部分有相同的前缀
 * 这样我们就可以枚举这个回文串的中心字符，然后求出这个位置和对称位置的lcp，这个的两倍就是
 * 回文串的长度，根据height数组的性质，lcp(i,j)=min(height[rank[i]+1],....,height[rank[j]])
 * 这个是区间最小值，用rmq或者线段树等都可以在O(n*logn)的复杂度内求出来，这样枚举每个位置，再取
 * 一个最大值就可以了，当然枚举的回文中心有奇数和偶数之分，都算一下就ok了，这样总的复杂度就是
 * O(n*n*logn)，就可以ac了。
 */

#include <bits/stdc++.h>
using namespace std;

const int MAXN = 2000+100;

int sa[MAXN];
int t1[MAXN],t2[MAXN],c[MAXN];
int Rank[MAXN],height[MAXN];
void build_sa(int s[],int n,int m)
{
    int i,j,p,*x=t1,*y=t2;
    for(i=0;i<m;i++)c[i]=0;
    for(i=0;i<n;i++)c[x[i]=s[i]]++;
    for(i=1;i<m;i++)c[i]+=c[i-1];
    for(i=n-1;i>=0;i--)sa[--c[x[i]]]=i;
    for(j=1;j<=n;j<<=1)
    {
        p=0;
        for(i=n-j;i<n;i++)y[p++]=i;
        for(i=0;i<n;i++)if(sa[i]>=j)y[p++]=sa[i]-j;
        for(i=0;i<m;i++)c[i]=0;
        for(i=0;i<n;i++)c[x[y[i]]]++;
        for(i=1;i<m;i++)c[i]+=c[i-1];
        for(i=n-1;i>=0;i--)sa[--c[x[y[i]]]]=y[i];
        swap(x,y);
        p=1;x[sa[0]]=0;
        for(i=1;i<n;i++)
            x[sa[i]]=y[sa[i-1]]==y[sa[i]] && y[sa[i-1]+j]==y[sa[i]+j]?p-1:p++;
        if(p>=n)break;
        m=p;
    }
}
void getHeight(int s[],int n)
{
    int i,j,k=0;
    for(i=0;i<=n;i++) Rank[sa[i]]=i;
    for(i=0;i<n;i++)
    {
        if(k)k--;
        j=sa[Rank[i]-1];
        while(s[i+k]==s[j+k])k++;
        height[Rank[i]]=k;
    }
}
struct SparseTable
{
    int rmq[MAXN];
    int mm[MAXN];
    int dp[MAXN][20];
    void init(int n)
    {
        mm[0]=-1;
        for(int i=1;i<=n;i++)
        {
            mm[i]=((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
            dp[i][0]=i;
        }
        for(int j=1;j<=mm[n];j++)
            for(int i=1;i+(1<<j)-1<=n;i++)
                dp[i][j]=rmq[dp[i][j-1]]<rmq[dp[i+(1<<(j-1))][j-1]]?dp[i][j-1]:dp[i+(1<<(j-1))][j-1];
    }
    int query(int a,int b)
    {
        if(a>b) swap(a,b);
        int k=mm[b-a+1];
        return rmq[dp[a][k]]<=rmq[dp[b-(1<<k)+1][k]]?dp[a][k]:dp[b-(1<<k)+1][k];
    }
}ST;
int lcp(int a,int b)
{
    a=Rank[a],b=Rank[b];
    if(a>b) swap(a,b);
    return height[ST.query(a+1,b)];
}
int ss[MAXN];
char str[MAXN];

int main()
{
    scanf("%s",str);
    int len=strlen(str);
    int n=2*len+1;
    for(int i=0;i<len;i++) ss[i]=str[i];
    ss[len]=1;
    for(int i=0;i<len;i++) ss[n-i-1]=str[i];
    ss[n]=0;
    build_sa(ss,n+1,128);
    getHeight(ss,n);
    for(int i=1;i<=n;i++) ST.rmq[i]=height[i];
    ST.init(n);
    int st,ans=0;
    for(int i=0;i<len;i++)
    {
        int tmp=lcp(i,n-i);
        if(2*tmp>ans)//偶对称
        {
            ans=2*tmp;
            st=i-tmp;
        }
        tmp=lcp(i,n-i-1);
        if(2*tmp-1>ans)//奇对称
        {
            ans=2*tmp-1;
            st=i-tmp+1;
        }
    }
    str[ans+st]=0;
    printf("%s\n",str+st);
    return 0;
}