spoj1812 && bzoj2946 公共串 [后缀自动机]

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Description

LCS2 - Longest Common Substring II

A string is finite sequence of characters over a non-empty finite set Σ.
In this problem, Σ is the set of lowercase letters.
Substring, also called factor, is a consecutive sequence of characters occurrences at least once in a string.
Now your task is a bit harder, for some given strings, find the length of the longest common substring of them.
Here common substring means a substring of two or more strings.


Input

The input contains at most 10 lines, each line consists of no more than 100000 lowercase letters, representing a string.


Output

The length of the longest common substring. If such string doesn't exist, print "0" instead.


Sample Input

alsdfkjfjkdsal
fdjskalajfkdsla
aaaajfaaaa


Sample Input

2


Solution

这道题是spoj1811的加强版,求多个串的最长公共子串,思路和上一题一样,对第一个输入的串建一个后缀自动机,其他的串在自动机上面匹配,匹配时记录每个位置能匹配到的最大值,每次更新ans为对应位置上最大值的最小值.需要注意的是当走到一个点时要更新它的所有父亲节点的maxs值,原因很简单,经过Tsort()后整个序列是按maxl由小到大排布的,这样可以看成他们的parent树由父亲到儿子节点的过程,这样逆序循环一遍,同时更新父亲节点的信息既可.


Code

#include<bits/stdc++.h>
#define maxn 200010
//#define DEBUG
using namespace std;
int n,answer=0;
int ans[maxn],tmp[maxn],sum[maxn],maxs[maxn];
char ch[maxn];

struct SAM
{
    int root,tot,last;
    int son[maxn][26],fa[maxn],maxl[maxn];
    
    void init()
    {
        root=tot=last=1;
        memset(son,0,sizeof(son));
        memset(fa,0,sizeof(fa));
        memset(maxl,0,sizeof(maxl));
    }
    
    void add(int pos)
    {
        int x=ch[pos]-'a';
        int np=++tot,p=last;
        last=np,maxl[np]=maxl[p]+1;
        for( ; p&&!son[p][x] ; p=fa[p] ) son[p][x]=np;
        if(!p) fa[np]=root;
        else
        {
            int q=son[p][x];
            if(maxl[q]==maxl[p]+1) fa[np]=q;
            else
            {
                int nq=++tot;
                maxl[nq]=maxl[p]+1;
                memcpy(son[nq],son[q],sizeof(son[q]));
                fa[nq]=fa[q];
                fa[q]=fa[np]=nq;
                for( ; son[p][x]==q ; p=fa[p] ) son[p][x]=nq;
            }
        }
    }
    
    void sort()
    {
        for(int i=1;i<=tot;++i) sum[maxl[i]]++;
        for(int i=1;i<=n;++i) sum[i]+=sum[i-1];
        for(int i=1;i<=tot;++i) tmp[sum[maxl[i]]--]=i;
    #ifdef DEBUG
        for (int i=1;i<=tot;i++) printf("tmp[%d]=%d dis[tmp[%d]]=%d\n",i,tmp[i],i,maxl[tmp[i]]);
    #endif 
    }
    
    void build()
    {
        init();
        scanf("%s",ch+1);
        n=strlen(ch+1);
        for(int i=1;i<=n;++i) add(i);
        for(int i=1;i<=tot;++i) ans[i]=maxl[i];
        sort();
    }
    
    void update()
    {
        for(int i=tot;i>0;i--)
        {
            int x=tmp[i];
            ans[x]=min(ans[x],maxs[x]);
            if(maxs[x]&&fa[x]) maxs[fa[x]]=maxl[fa[x]];
        }
    }
    
    void work()
    {
        memset(maxs,0,sizeof(maxs));
        int len=0,p=root;
        for(int i=1;i<=n;++i)
        {
            int x=ch[i]-'a';
            if(son[p][x]) len++,p=son[p][x];
            else
            {
                for( ; p&&!son[p][x] ; p=fa[p] ) ;
                if(!p) len=0,p=root;
                else len=maxl[p]+1,p=son[p][x];
            }
            maxs[p]=max(maxs[p],len);
        }
        update();
    }
    
} sam ;

int main()
{
    memset(ans,0,sizeof(ans));
    sam.build();
    while(scanf("%s",ch+1)==1)
    {
        n=strlen(ch+1);
        sam.work();
    }
#ifdef DEBUG
    for(int i=1;i<=sam.tot;++i) printf("ans[%d]=%d\n",i,ans[i]);
#else
    for(int i=1;i<=sam.tot;++i) answer=max(answer,ans[i]);
#endif
    printf("%d",answer);
} 

/*
alsdfkjfjkdsal
fdjskalajfkdsla
aaaajfaaaa
--------------
2
*/

参考资料

http://blog.csdn.net/thy_asdf/article/details/51569443


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