【SPOJ1811】Longest Common Substring（后缀自动机）

题意：给定两个仅含小写字母的字符串，求他们最长公共子串的长度

n<=250000

思路：

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef pair<ll,ll> Pll;
 typedef vector<int> VI;
 typedef vector<PII> VII;
 typedef pair<ll,int>P;
 #define N  510000
 #define M  151000
 #define fi first
 #define se second
 #define MP make_pair
 #define pi acos(-1)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 #define lowbit(x) x&(-x)
 #define Rand (rand()*(1<<16)+rand())
 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 #define ls p<<1
 #define rs p<<1|1

 const int MOD=,inv2=(MOD+)/;
       double eps=1e-;
       ll INF=1e18;
       ll inf=5e13;
       int dx[]={-,,,};
       int dy[]={,,-,};

 char s[N];
 int n,i,x,p,q,np,nq,cnt,L,st[N],c[N][],f[N],pos[N],bl[N],to[N],b[N],sz[N];

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 void add(int x)
 {
     p=np;
     st[np=++cnt]=st[p]+;
     to[np]=i;
     pos[i]=np;
     for(;p&&!c[p][x];p=f[p]) c[p][x]=np;
     if(!p) f[np]=;
      else if(st[p]+==st[q=c[p][x]]) f[np]=q;
       else
       {
           st[nq=++cnt]=st[p]+;
           memcpy(c[nq],c[q],sizeof c[q]);
           f[nq]=f[q];
           f[q]=f[np]=nq;
           for(;p&&c[p][x]==q;p=f[p]) c[p][x]=nq;
       }
 }

 int main()
 {
     //freopen("1.in","r",stdin);
     //freopen("1.out","w",stdout);
     np=cnt=;
     scanf("%s",s);
     n=strlen(s);
     rep(i,,n-) add(s[i]-'a');

     rep(i,,cnt) b[st[i]]++;
     rep(i,,n) b[i]+=b[i-];
     rep(i,,cnt) bl[b[st[i]]--]=i;
     for(i=,p=;i<n;i++) sz[p=c[p][s[i]-'a']]++;
     for(i=cnt;i;i--) sz[f[bl[i]]]+=sz[bl[i]];
     //rep(i,0,n-1) printf("%d ",sz[pos[i]]);
     scanf("%s",s);
     n=strlen(s);
     int ans=;
     for(p=,L=i=;i<n;i++)
     {
         if(c[p][x=s[i]-'a']) p=c[p][x],L++;
          else
          {
               for(;!c[p][x]&&p;p=f[p]);
                if(!p) p=,L=;
                 else L=st[p]+,p=c[p][x];
          }
         ans=max(ans,L);
     }
     printf("%d\n",ans);
     return ;
 }