spoj LCS 【后缀自动机】

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琦不会后缀自动机……
是以前太浪了……
所以所有东西都留到了noi前来学……
马上狗牌退役了TAT（心塞qwq

题目大意：给出两个串A,B，求A、B的最长公共子串

对A建后缀自动机，然后用B去匹配，若能匹配上就转移到儿子，否则沿着parent树向上跳

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>#define N 500005using namespace std;int n,m,last = 1,tot = 1,p,q,np,nq,ans;int son[N][26],par[N],mx[N];char a[N],b[N];int new_node(int x){    mx[++ tot] = x;    return tot;}void add(int x){    p = last;    np = new_node(mx[p] + 1);    for (;p && !son[p][x];p = par[p]) son[p][x] = np;    if (!p) par[np] = 1;    else    {        q = son[p][x];        if (mx[q] == mx[p] + 1) par[np] = q;        else        {            nq = new_node(mx[p] + 1);            memcpy(son[nq],son[q],sizeof(son[nq]));            par[nq] = par[q],par[q] = par[np] = nq;            for (;son[p][x] == q;p = par[p]) son[p][x] = nq;        }    }    last = np;}int main(){    scanf("%s%s",a + 1,b + 1);    n = strlen(a + 1),m = strlen(b + 1);    for (int i = 1;i <= n;i ++) add(a[i] - 'a');    for (int i = 1,p = 1,l = 0;i <= m;i ++)    {        int x = b[i] - 'a';        if (son[p][x]) l ++,p = son[p][x];        else        {            for (;p && !son[p][x];p = par[p]);            if (!p) l = 0,p = 1;            else l = mx[p] + 1,p = son[p][x];        }        ans = max(ans,l);    }    cout << ans << endl;    return 0;}

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