【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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*/