poj 1743 Musical Theme【后缀自动机】

题目大意：给定一个串，求最长的两个子串，满足互不相交且一个串可以通过整体加上另一个数变成另一个串

先差分，然后就是找两个相同的不相交的串，考虑对于每个点，记录它的right集的最值，差值就是子串长度。取个max就好

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#define N 40005
#define INF 1000000000
using namespace std;

int n,m,last,tot,p,q,np,nq,ans;
int l
,r
,P
,sm
;;
int son
[180],par
,mx
;
int s
;

int new_node(int x)
{
mx[++ tot] = x;
return tot;
}

void add(int x)
{
p = last;
np = new_node(mx[p] + 1);l[np] = r[np] = mx[np];
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);l[nq] = 1e9,r[nq] = 0;
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()
{
while (~scanf("%d",&n) && n)
{
memset(son,0,sizeof(son));last = tot = 1;
memset(sm,0,sizeof(sm));ans = 0;
for (int i = 1;i <= n;i ++) scanf("%d",&s[i]);
for (int i = 1;i < n;i ++) s[i] = s[i + 1] - s[i] + 88;n --;
for (int i = 1;i <= n;i ++) add(s[i]);
for (int i = 1;i <= tot;i ++) sm[mx[i]] ++;
for (int i = 1;i <= n;i ++) sm[i] += sm[i - 1];
for (int i = 1;i <= tot;i ++) P[sm[mx[i]] --] = i;
for (int i = tot;i > 1;i --)
{
int x = P[i];
l[par[x]] = min(l[par[x]],l[x]);
r[par[x]] = max(r[par[x]],r[x]);
}
for (int i = 2;i <= tot;i ++)
ans = max(ans,min(r[i] - l[i],mx[i]));
cout << (ans < 4 ? 0 : ans + 1) << endl;
}

return 0;
}