SPOJ-LCS Longest Common Substring 【后缀自动机】

题目分析：

用没出现过的字符搞拼接。搞出right树，找right集合的最小和最大。如果最小和最大分居两侧可以更新答案。

代码：

#include<bits/stdc++.h>
using namespace std;

const int maxn = 1000500;

int son[maxn][30],fa[maxn],maxlen[maxn],root,num,sigma = 27;
int minn[maxn],maxx[maxn];// lright rright

string str,ss;
int n,m;
vector<int> T[maxn];

int addnew(int dt){maxlen[++num] = dt;return num;}
int ins(char x,int p,int hhhh){
    int np = addnew(maxlen[p]+1); minn[np] = maxx[np] = hhhh;
    while(p && !son[p][x-'a']) son[p][x-'a'] = np,p = fa[p];
    if(!p){fa[np] = root; return np;}
    else{
    int q = son[p][x-'a'];
    if(maxlen[q]==maxlen[p]+1){fa[np]=q;return np;}
    else{
        int nq = addnew(maxlen[p]+1);
        for(int i=0;i<sigma;i++) son[nq][i] = son[q][i];
        fa[nq] = fa[q]; fa[q] = fa[np] = nq;
        while(p && son[p][x-'a'] == q) son[p][x-'a'] = nq,p = fa[p];
        return np;
    }
    }
}

void dfs(int now){
    if(!minn[now]) minn[now] = 1e9;
    for(int i=0;i<T[now].size();i++){
    dfs(T[now][i]);
    minn[now] = min(minn[now],minn[T[now][i]]);
    maxx[now] = max(maxx[now],maxx[T[now][i]]);
    }
}

void work(){
    int lst = addnew(0);root = lst;
    for(int i=0;i<str.length();i++)
    lst = ins(str[i],lst,i+1);
    for(int i=2;i<=num;i++) T[fa[i]].push_back(i);
    dfs(1); int ans = 0;
    for(int i=1;i<=num;i++)if(minn[i]<=n&&maxx[i]>n+1)ans=max(ans,maxlen[i]);
    cout<<ans<<endl;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> str; cin >> ss;
    n = str.length(); m = ss.length();
    str += ('z'+1); str += ss;
    work();
    return 0;
}