SPOJ LCS2 - Longest Common Substring II 后缀自动机多个串的LCS

LCS2 - Longest Common Substring II

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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.

Example

Input:

alsdfkjfjkdsal

fdjskalajfkdsla

aaaajfaaaa

Output:

2

Notice: new testcases added

题意：

　　求多个串的最长公共字串

题解：

　　将第一个串建立后缀自动机

　　根据拓扑排序更新每个状态节点下所能向前延伸的长度，每次匹配一个串都更新即可

　　是不是很模糊

#include <bits/stdc++.h>

inline long long read(){long long x=,f=;char ch=getchar();while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}return x*f;}

using namespace std;

const int N = 3e5+;

const long long mod = ;

int isPlus[N * ],endpos[N * ];int d[N * ];

int tot,slink[*N],trans[*N][],minlen[*N],maxlen[*N],pre;

int newstate(int _maxlen,int _minlen,int* _trans,int _slink){

    maxlen[++tot]=_maxlen;minlen[tot]=_minlen;

    slink[tot]=_slink;

    if(_trans)for(int i=;i<;i++)trans[tot][i]=_trans[i],d[_trans[i]]+=;

    return tot;

}

int add_char(char ch,int u){

    int c=ch-'a',v=u;

    int z=newstate(maxlen[u]+,-,NULL,);

    isPlus[z] = ;

    while(v&&!trans[v][c]){trans[v][c]=z;d[z]+=;v=slink[v];}

    if(!v){ minlen[z]=;slink[z]=;return z;}

    int x=trans[v][c];

    if(maxlen[v]+==maxlen[x]){slink[z]=x;minlen[z]=maxlen[x]+;return z;}

    int y=newstate(maxlen[v]+,-,trans[x],slink[x]);

    slink[z]=slink[x]=y;minlen[x]=minlen[z]=maxlen[y]+;

    while(v&&trans[v][c]==x){trans[v][c]=y;d[x]--,d[y]++;v=slink[v];}

    minlen[y]=maxlen[slink[y]]+;

    return z;

}

void init_sam() {

    for(int i = ; i <= tot; ++i)

        for(int j = ; j < ; ++j) trans[i][j] = ;

    pre = tot = ;

}

int dp[N],ans,n,vis[N],cnt[N],pos[N];

char a[N];

int main() {

    int f = ;

    while(scanf("%s",a)!=EOF) {

        n = strlen(a);

        if(f) {

            init_sam();

            for(int i = ; i < n; ++i)

                pre = add_char(a[i],pre);

            f = ;

            for(int i = ; i <= n; ++i)cnt[i] = ;

            for(int i = ; i <= tot; ++i)cnt[maxlen[i]] += ;

            for(int i = ; i <= n; ++i)cnt[i]+=cnt[i-];

            for(int i = tot; i >= ; --i)pos[cnt[maxlen[i]]--] = i;

        }

        else

        {

            int now = ,sum = ;

            for(int i = ; i < n; ++i)

            {

                if(trans[now][a[i]-'a']) {

                    sum++;

                    now = trans[now][a[i]-'a'];

                    dp[now] = max(dp[now],sum);

                }

                else {

                    while(now) {

                        now = slink[now];

                        if(trans[now][a[i]-'a']) break;

                    }

                    if(!now) now = ,sum = ;

                    else

                    {

                        sum = maxlen[now] + ;

                        now = trans[now][a[i]-'a'];

                        dp[now] = max(dp[now],sum);

                    }

                }

            }

            for(int i = tot; i >= ; --i) {

                int v = pos[i];

                maxlen[v] = min(maxlen[v],dp[v]);

                dp[slink[v]] = max(dp[slink[v]],dp[v]);

                dp[v] = ;

            }

        }

    }

    int ans = ;

    for(int i = ; i <= tot; ++i) {

        ans = max(ans,maxlen[i]);

    }

    printf("%d\n",ans);

    return ;

}

/*

aabbcd

babbefg

cdbbabb

*/