URAL 1517 Freedom of Choice

URAL第一A。。。。后缀数组第一题。。。。。
大白书上关于后缀数组的部分好多bug。。。。。。
总算被我一一坑过去了。。。。。
搞会了后缀数组的正确姿势。。。。

Freedom of Choice Time Limit: 2000MS   Memory Limit: 65536KB   64bit IO Format: %I64d & %I64u [Submit]   [Go Back]   [Status]   Description 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  Ncapital 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 input output 28 VOTEFORTHEGREATALBANIAFORYOU CHOOSETHEGREATALBANIANFUTURE THEGREATALBANIA Source Problem Author: Ilya Grebnov, Nikita Rybak, Dmitry Kovalioff  Problem Source: Timus Top Coders: Third Challenge  [Submit]   [Go Back]   [Status]

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int N=4e5+10;

char s[N];
int sa[N],rank[N],rank2[N],height[N],c[N],*x,*y;

void radix_sort(int n,int sz)
{
    memset(c,0,sizeof(c));
    for(int i=0;i<n;i++) c[x[y[i]]]++;
    for(int i=1;i<sz;i++) c[i]+=c[i-1];
    for(int i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];
}

void get_sa(char* s,int n,int sz=128)
{
    x=rank;y=rank2;
    for(int i=0;i<n;i++)
        x[i]=s[i],y[i]=i;
    radix_sort(n,sz);
    for(int len=1;len<n;len<<=1)
    {
        int yid=0;
        for(int i=n-len;i<n;i++) y[yid++]=i;
        for(int i=0;i<n;i++) if(sa[i]>=len) y[yid++]=sa[i]-len;

        radix_sort(n,sz);

        swap(x,y);
        x[sa[0]]=yid=0;
        for(int i=1;i<n;i++)
        {
            if(y[sa[i-1]]==y[sa[i]]&&sa[i-1]+len<n&&sa[i]+len<n
               &&y[sa[i-1]+len]==y[sa[i]+len])
                x[sa[i]]=yid;
            else x[sa[i]]=++yid;
        }
        sz=yid+1;
        if(sz>=n) break;
    }
    for(int i=0;i<n;i++)
        rank[i]=x[i];
}

void get_height(char* s,int n)
{
    int k=0;
    for(int i=0;i<n;i++)
    {
        if(rank[i]==0) continue;
        k=max(0,k-1);
        int j=sa[rank[i]-1];
        while(s[i+k]==s[j+k]) k++;
        height[rank[i]]=k;
    }
}

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        scanf("%s",s);
        s[n]=127;
        scanf("%s",s+n+1);
        int nn=strlen(s);
        get_sa(s,nn);
        get_height(s,nn);
        int be=0,len=0;
        for(int i=1;i<nn;i++)
        {
            if((sa[i]<n&&sa[i-1]>n)||(sa[i]>n&&sa[i-1]<n))
            {
                if(height[i]>len)
                {
                    be=sa[i]; len=height[i];
                }
            }
        }
        for(int i=be;i<be+len;i++) putchar(s[i]);
        putchar(10);
    }
    return 0;
}