URAL 1297 最長迴文子串(後綴數組)

1297. Palindrome

Time limit: 1.0 second
Memory limit: 64 MB

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

output

ArozaupalanalapuazorA

/*
URAL 1297 最長迴文子串(後綴數組)

算法合集之《後綴數組——處理字符串的有力工具》：
窮舉每一位，然後計算以這個字符爲中心的最長迴文子串。注意這裏要分兩
種情況，一是迴文子串的長度爲奇數，二是長度爲偶數。兩種情況都可以轉化爲
求一個後綴和一個反過來寫的後綴的最長公共前綴。具體的做法是：將整個字符
串反過來寫在原字符串後面，中間用一個特殊的字符隔開。這樣就把問題變爲了
求這個新的字符串的某兩個後綴的最長公共前綴。

所以我們只需先對初始的字符串進行一下處理，然後分別進行奇偶判斷得到
最長迴文子串的位置和長度

hhh-2016-03-13 15:41:30
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
#include <map>
using namespace std;
typedef long long ll;
typedef long double ld;
#define lson (i<<1)
#define rson ((i<<1)|1)
const int maxn = 5005;

int t1[maxn],t2[maxn],c[maxn];
bool cmp(int *r,int a,int b,int l)
{
    return r[a]==r[b] &&r[l+a] == r[l+b];
}

void get_sa(int str[],int sa[],int Rank[],int height[],int n,int m)
{
    n++;
    int p,*x=t1,*y=t2;
    for(int i = 0; i < m; i++) c[i] = 0;
    for(int i = 0; i < n; i++) c[x[i] = str[i]]++;
    for(int i = 1; i < m; i++) c[i] += c[i-1];
    for(int i = n-1; i>=0; i--) sa[--c[x[i]]] = i;
    for(int j = 1; j <= n; j <<= 1)
    {
        p = 0;
        for(int i = n-j; i < n; i++) y[p++] = i;
        for(int i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i]-j;
        for(int i = 0; i < m; i++) c[i] = 0;
        for(int i = 0; i < n; i++) c[x[y[i]]]++ ;
        for(int i = 1; i < m; i++) c[i] += c[i-1];
        for(int i = n-1; i >= 0; i--)  sa[--c[x[y[i]]]] = y[i];

        swap(x,y);
        p = 1;
        x[sa[0]] = 0;
        for(int i = 1; i < n; i++)
            x[sa[i]] = cmp(y,sa[i-1],sa[i],j)? p-1:p++;
        if(p >= n) break;
        m = p;
    }
    int k = 0;
    n--;
    for(int i = 0; i <= n; i++)
        Rank[sa[i]] = i;
    for(int i = 0; i < n; i++)
    {
        if(k) k--;
        int j = sa[Rank[i]-1];
        while(str[i+k] == str[j+k]) k++;
        height[Rank[i]] = k;
    }
}


int mm[maxn];
int dp[20][maxn];
int Rank[maxn],height[maxn];
int sa[maxn],str[maxn];
char ts[maxn];

void ini_RMQ(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]);

    for(int i =1;i <= n;i++)
        dp[0][i] = height[i];
    for(int i = 1;i <= mm[n];i++)
    {
        for(int j = 1;j+(1<<i)-1 <= n;j++)
        {
            int a = dp[i-1][j];
            int b = dp[i-1][j+(1<<(i-1))];
            dp[i][j] = min(a,b);
        }
    }
}

int askRMQ(int a,int b)
{
    int t = mm[b-a+1];
    b -= (1<<t)-1;
    return min(dp[t][a],dp[t][b]);
}


int fin(int a,int b)
{
    a = Rank[a],b = Rank[b];
    if(a > b) swap(a,b);
    return askRMQ(a+1,b);
}

int main()
{
    while(scanf("%s",ts) != EOF)
    {
        int len = strlen(ts);
        for(int i = 0;i < len;i++)
            str[i] = ts[i];
        str[len] = 1;
        for(int i = 0;i < len;i++)
            str[i+len+1] = ts[len-i-1];
        str[len*2+1] = 0;

        get_sa(str,sa,Rank,height,2*len+1,128);
        ini_RMQ(2*len+1);

        int ans = 0,pos;
        int tp;
        for(int i = 0;i < len;i++)
        {
            tp = fin(i,len*2+1-i);
            if(tp*2 > ans)
            {
                ans = tp*2;
                pos = i-tp;
            }
            tp = fin(i,len*2-i);
            if(tp*2-1 > ans)
            {
                ans = tp*2-1;
                pos = i-tp+1;
            }
        }
        ts[pos+ans] = 0;
        printf("%s\n",ts+pos);
    }
    return 0;
}