bzoj3926: [Zjoi2015]诸神眷顾的幻想乡 后缀自动机在tire树上拓展

题意:有棵树每个点有个颜色(不超过10种),每个节点不超过20个儿子,问你每两点之间的颜色序列不同的有多少种
题解:先建出树,对于每个叶子节点,bfs一遍建在sam上,每次保留当前点在sam上的位置,拓展时用父亲节点在sam上的位置当成last即可.然后统计sam本质不同的字符串有多少个
注:dfs建树复杂度是错的,但是也能过这题

/**************************************************************
    Problem: 3926
    User: walfy
    Language: C++
    Result: Accepted
    Time:3612 ms
    Memory:270836 kb
****************************************************************/

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
//#define C 0.5772156649
//#define ls l,m,rt<<1
//#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define ull unsigned long long
//#define base 1000000000000000000
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const ull ba=233;
const db eps=1e-5;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=100000+10,maxn=1000000+10,inf=0x3f3f3f3f;

struct SAM{
    int last,cnt;
    int ch[N*40][15],fa[N*40],l[N*40];
    int ins(int y,int x)
    {
        last=y;
        int p=last,np=++cnt;last=np;l[np]=l[p]+1;
        for(;p&&!ch[p][x];p=fa[p])ch[p][x]=np;
        if(!p)fa[np]=1;
        else
        {
            int q=ch[p][x];
            if(l[q]==l[p]+1)fa[np]=q;
            else
            {
                int nq=++cnt;l[nq]=l[p]+1;
                memcpy(ch[nq],ch[q],sizeof ch[q]);
                fa[nq]=fa[q];fa[q]=fa[np]=nq;
                for(;ch[p][x]==q;p=fa[p])ch[p][x]=nq;
            }
        }
        return last;
    }
    void build()
    {
        cnt=last=1;
    }
    ll cal()
    {
        ll ans=0;
        for(int i=1;i<=cnt;i++)ans+=l[i]-l[fa[i]];
        return ans;
    }
}sam;
int a[N],d[N];
bool vis[N];
struct Tire{
    vector<int>v[N];
    int id[N];
    void ins(int a,int b){v[a].pb(b);v[b].pb(a);}
    void bfs(int u)
    {
        memset(vis,0,sizeof vis);
        queue<int>q;q.push(u);vis[u]=1;
        id[u]=sam.ins(1,a[u]);
        while(!q.empty())
        {
            u=q.front();q.pop();
            for(int i=0;i<v[u].size();i++)if(!vis[v[u][i]])
            {
                int x=v[u][i];
                vis[x]=1;q.push(x);
                id[x]=sam.ins(id[u],a[x]);
            }
        }
    }
}t;
int main()
{
    int n,c;scanf("%d%d",&n,&c);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    for(int i=1;i<n;i++)
    {
        int a,b;scanf("%d%d",&a,&b);
        t.ins(a,b);d[a]++,d[b]++;
    }
    sam.build();
    for(int i=1;i<=n;i++)if(d[i]==1)t.bfs(i);
    printf("%lld\n",sam.cal());
    return 0;
}
/********************

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原文地址：https://www.cnblogs.com/acjiumeng/p/10661531.html