URAL 1297 Palindrome（后缀数组求最长回文子串）

 Palindrome

Time Limit:1000MS     Memory Limit:16384KB     64bit IO Format:%I64d & %I64u

Submit Status

Appoint description:  System Crawler  (2013-01-30)

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

input

ThesampletextthatcouldbereadedthesameinbothordersArozaupalanalapuazorA

output

ArozaupalanalapuazorA

 

 

 

 

作为后缀数组的入门题，但是求最长回文子串有更加快速的算法。

//============================================================================
// Name        : URAL.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
using namespace std;
const int MAXN=2010;

/*
*suffix array
*倍增算法  O(n*logn)
*待排序数组长度为n,放在0~n-1中，在最后面补一个0
*build_sa( ,n+1, );//注意是n+1;
*getHeight(,n);
*例如：
*n   = 8;
*num[]   = { 1, 1, 2, 1, 1, 1, 1, 2, $ };注意num最后一位为0，其他大于0
*rank[]  = { 4, 6, 8, 1, 2, 3, 5, 7, 0 };rank[0~n-1]为有效值，rank[n]必定为0无效值
*sa[]    = { 8, 3, 4, 5, 0, 6, 1, 7, 2 };sa[1~n]为有效值，sa[0]必定为n是无效值
*height[]= { 0, 0, 3, 2, 3, 1, 2, 0, 1 };height[2~n]为有效值
*
*/

int sa[MAXN];//SA数组，表示将S的n个后缀从小到大排序后把排好序的
             //的后缀的开头位置顺次放入SA中
int t1[MAXN],t2[MAXN],c[MAXN];//求SA数组需要的中间变量，不需要赋值
int rank[MAXN],height[MAXN];
//待排序的字符串放在s数组中，从s[0]到s[n-1],长度为n,且最大值小于m,
//除s[n-1]外的所有s[i]都大于0，r[n-1]=0
//函数结束以后结果放在sa数组中
void build_sa(int s[],int n,int m)
{
    int i,j,p,*x=t1,*y=t2;
    //第一轮基数排序，如果s的最大值很大，可改为快速排序
    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;
        //直接利用sa数组排序第二关键字
        for(i=n-j;i<n;i++)y[p++]=i;//后面的j个数第二关键字为空的最小
        for(i=0;i<n;i++)if(sa[i]>=j)y[p++]=sa[i]-j;
        //这样数组y保存的就是按照第二关键字排序的结果
        //基数排序第一关键字
        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];
        //根据sa和x数组计算新的x数组
        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;//下次基数排序的最大值
    }
}
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;
    }
}

int RMQ[MAXN];
int mm[MAXN];
int best[20][MAXN];
void initRMQ(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++)best[0][i]=i;
    for(int i=1;i<=mm[n];i++)
        for(int j=1;j+(1<<i)-1<=n;j++)
        {
            int a=best[i-1][j];
            int b=best[i-1][j+(1<<(i-1))];
            if(RMQ[a]<RMQ[b])best[i][j]=a;
            else best[i][j]=b;
        }
}
int askRMQ(int a,int b)
{
    int t;
    t=mm[b-a+1];
    b-=(1<<t)-1;
    a=best[t][a];b=best[t][b];
    return RMQ[a]<RMQ[b]?a:b;
}

int lcp(int a,int b)
{
    a=rank[a];b=rank[b];
    if(a>b)swap(a,b);
    return height[askRMQ(a+1,b)];
}
int r[MAXN];
char str[MAXN];
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    while(scanf("%s",&str)==1)
    {
        int len=strlen(str);
        int n=2*len+1;
        for(int i=0;i<len;i++)r[i]=str[i];
        r[len]=1;//这个和其他的都应该不同
        for(int i=0;i<len;i++)r[i+len+1]=str[len-1-i];
        r[n]=0;
        build_sa(r,n+1,128);
        getHeight(r,n);
        for(int i=1;i<=n;i++)RMQ[i]=height[i];
        initRMQ(n);
        int ans=0,st;
        int tmp;
        for(int i=0;i<len;i++)
        {
            tmp=lcp(i,n-i);//偶对称
            if(2*tmp>ans)
            {
                ans=2*tmp;
                st=i-tmp;
            }
            tmp=lcp(i,n-i-1);//奇数对称
            if(2*tmp-1>ans)
            {
                ans=2*tmp-1;
                st=i-tmp+1;
            }
        }
        str[st+ans]=0;
        printf("%s\n",str+st);
    }
    return 0;
}