URAL1517-Freedom of Choice

Freedom of Choice

Time limit: 2.0 second
Memory limit: 64 MB

Background

Before Albanian people could bear with the freedom of speech (this story is fully described in the problem  "Freedom of speech"), another freedom - the freedom of choice - came down on them. In the near future, the inhabitants will have to face the first democratic Presidential election in the history of their country.

Outstanding Albanian politicians liberal Mohammed Tahir-ogly and his old rival conservative Ahmed Kasym-bey declared their intention to compete for the high post.

Problem

According to democratic traditions, both candidates entertain with digging dirt upon each other to the cheers of their voters' approval. When occasion offers, each candidate makes an election speech, which is devoted to blaming his opponent for corruption, disrespect for the elders and terrorism affiliation. As a result the speeches of Mohammed and Ahmed have become nearly the same, and now it does not matter for the voters for whom to vote.

The third candidate, a chairman of Albanian socialist party comrade Ktulhu wants to make use of this situation. He has been lazy to write his own election speech, but noticed, that some fragments of the speeches of Mr. Tahir-ogly and Mr. Kasym-bey are completely identical. Then Mr. Ktulhu decided to take the longest identical fragment and use it as his election speech.

Input

The first line contains the integer number  N (1 ≤  N ≤ 100000). The second line contains the speech of Mr. Tahir-ogly. The third line contains the speech of Mr. Kasym-bey. Each speech consists of  N capital latin letters.

Output

You should output the speech of Mr. Ktulhu. If the problem has several solutions, you should output any of them.

Sample

input	output
28 VOTEFORTHEGREATALBANIAFORYOU CHOOSETHEGREATALBANIANFUTURE	THEGREATALBANIA

Problem Author: Ilya Grebnov, Nikita Rybak, Dmitry Kovalioff
Problem Source: Timus Top Coders: Third Challenge

题意：给你两个字符串，求出它们的最长公共子串

解题思路：后缀数组，将两个字符串连接起来，中间用没出现过的字符连接即可

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <map>

using namespace std;

#define N 200010

struct Sa
{
    char s[N],s1[N],s2[N];
    int rk[2][N],sa[N],h[N],w[N],now,n;
    int rmq[N][20],lg[N];

    bool GetS()
    {
        if(scanf("%d",&n)==EOF) return false;
        scanf("%s",s1+1);
        scanf("%s",s2+1);
        for(int i=1;i<=n;i++) s[i]=s1[i];
        s[n+1]='$';
        for(int i=n+2;i<=n+n+2;i++) s[i]=s2[i-n-1];
        n=n+n+1;
        return true;
    }

    void getsa(int z,int &m)
    {
        int x=now,y=now^=1;
        for(int i=1;i<=z;i++) rk[y][i]=n-i+1;
        for(int i=1,j=z;i<=n;i++)
            if(sa[i]>z) rk[y][++j]=sa[i]-z;
        for(int i=1;i<=m;i++) w[i]=0;
        for(int i=1;i<=n;i++) w[rk[x][rk[y][i]]]++;
        for(int i=1;i<=m;i++) w[i]+=w[i-1];
        for(int i=n;i>=1;i--) sa[w[rk[x][rk[y][i]]]--]=rk[y][i];
        for(int i=m=1;i<=n;i++)
        {
            int *a=rk[x]+sa[i],*b=rk[x]+sa[i-1];
            rk[y][sa[i]]=*a==*b&&*(a+z)==*(b+z)?m-1:m++;
        }
    }

    void getsa(int m)
    {
        now=rk[1][0]=sa[0]=s[0]=0;
        for(int i=1;i<=m;i++) w[i]=0;
        for(int i=1;i<=n;i++) w[s[i]]++;
        for(int i=1;i<=m;i++) rk[1][i]=rk[1][i-1]+(bool)w[i];
        for(int i=1;i<=m;i++) w[i]+=w[i-1];
        for(int i=1;i<=n;i++) rk[0][i]=rk[1][s[i]];
        for(int i=1;i<=n;i++) sa[w[s[i]]--]=i;
        rk[1][n+1]=rk[0][n+1]=0;
        for(int x=1,y=rk[1][m];x<=n&&y<=n;x<<=1) getsa(x,y);
        for(int i=1,j=0;i<=n;h[rk[now][i++]]=j?j--:j)
        {
            if(rk[now][i]==1) continue;
            int k=n-max(sa[rk[now][i]-1],i);
            while(j<=k&&s[sa[rk[now][i]-1]+j]==s[i+j]) ++j;
        }

    }

    void getrmq()
    {
        h[n+1]=h[1]=lg[1]=0;
        for(int i=2;i<=n;i++)
            rmq[i][0]=h[i],lg[i]=lg[i>>1]+1;
        for(int i=1;(1<<i)<=n;i++)
        {
            for(int j=2;j<=n;j++)
            {
                if(j+(1<<i)>n+1) break;
                rmq[j][i]=min(rmq[j][i-1],rmq[j+(1<<i-1)][i-1]);
            }
        }
    }

    int lcp(int x,int y)
    {
        int l=min(rk[now][x],rk[now][y])+1,r=max(rk[now][x],rk[now][y]);
        return min(rmq[l][lg[r-l+1]],rmq[r-(1<<lg[r-l+1])+1][lg[r-l+1]]);
    }

    void work()
    {
        int ans=-1;
        int mi;
        for(int i=2;i<=n;i++)
        {
            int MAX=max(sa[i-1],sa[i]);
            int MIN=min(sa[i-1],sa[i]);
            if(MAX>n/2&&MIN<n/2&&ans<h[i])
                ans=h[i],mi=MIN;
        }
        for(int i=mi;i<mi+ans;i++)
            printf("%c",s[i]);
        printf("\n");
    }
}sa;

int main()
{
    while(sa.GetS())
    {
        sa.getsa(300);
        sa.work();
    }
    return 0;
}