【SPOJ】Longest Common Substring II

【SPOJ】Longest Common Substring II

多个字符串求最长公共子串
还是将一个子串建SAM，其他字符串全部跑一边，记录每个点的最大贡献
由于是所有串，要对每个点每个字符串跑完后去最小值才是每个点的最终贡献

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<queue>
using namespace std;
typedef long long LL;
const LL maxn=200000;
LL nod,last,n,T;
LL len[maxn],fail[maxn],son[maxn][26],Ans[maxn],sum[maxn],c[maxn],p[maxn];
char s[maxn];
inline void Insert(LL c){
    LL np=++nod,p=last;
    len[np]=len[p]+1;
    last=np;
    while(p&&!son[p][c]){
        son[p][c]=np,
        p=fail[p];
    }
    if(!p)
        fail[np]=1;
    else{
        LL q=son[p][c];
        if(len[q]==len[p]+1)
            fail[np]=q;
        else{
            LL nq=++nod;
            len[nq]=len[p]+1;
            fail[nq]=fail[q];
            memcpy(son[nq],son[q],sizeof(son[q]));
            fail[np]=fail[q]=nq;
            while(p&&son[p][c]==q){
                son[p][c]=nq,
                p=fail[p];
            }
        }
    }
}
int main(){ 
    nod=last=1;
    scanf("%s",s);
    LL Len=strlen(s);
    for(LL i=0;i<Len;++i)
        Insert(s[i]-'a');
    for(LL i=1;i<=nod;++i)
        c[len[i]]++;
    for(int i=1;i<=nod;++i)
        c[i]+=c[i-1];
    for(int i=1;i<=nod;++i)
        p[c[len[i]]--]=i;
    for(int i=1;i<=nod;++i)
        Ans[i]=len[i];
    while(scanf("%s",s)!=EOF){
        memset(sum,0,sizeof(sum));
        LL now=1,cnt=0;
        Len=strlen(s);
        for(LL i=0;i<Len;++i){
            LL c=s[i]-'a';
            if(son[now][c])
                ++cnt,
                now=son[now][c];
            else{
                while(now&&!son[now][c])
                    now=fail[now];
                if(!now)
                    cnt=0,
                    now=1;
                else
                    cnt=len[now]+1,
                    now=son[now][c];
            }
            sum[now]=max(sum[now],cnt);
        }
        for(LL i=nod;i>=1;--i)
            sum[fail[p[i]]]=max(sum[fail[p[i]]],sum[p[i]]);
        for(int i=1;i<=nod;++i)
            Ans[i]=min(Ans[i],sum[i]);
    }
    LL ans=0;
    for(LL i=1;i<=nod;++i)
        ans=max(ans,Ans[i]);
    printf("%lld",ans);
    return 0;
}/*
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*/