ICPC2019徐州 L. Loli, Yen-Jen, and a cool problem

鏈接

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題解

這題就是在廣義後綴自動機的$parent$樹上跑倍增的裸題

建$trie$樹的時候要注意處理重邊

然後真沒啥好說的了，因爲實在是太裸了

代碼

#include <bits/stdc++.h>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 600010
#define maxe 600010
#define maxk 20
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
char s[maxn];
ll tb[maxn];
struct Trie
{
    int tot, ch[maxn][26], end[maxn];
    void init(){while(tot--)cl(ch[tot+1]),end[tot+1]=0;tot=1;}
    int* operator[](int index){return ch[index];}
    int append(int pos, int c)
    {
        int &t=ch[pos][c];
        t = t?t:t=++tot;
        end[t]++;
        return t;
    }
}trie;
struct EXSAM
{
    int tot, ch[maxn][26], fa[maxn], len[maxn], pref[maxn], tb[maxn];
    int* operator[](int u){return ch[u];}
    void init()
    {
        int i;
        rep(i,1,tot)cl(ch[i]),fa[i]=len[i]=pref[i]=0;
        tot=1;
    }
    int append(int las, int c, int cnt)
    {
        int p(las);
        len[las=++tot]=len[p]+1;
        pref[las]=cnt;
        for(;p and !ch[p][c];p=fa[p])ch[p][c]=las;
        if(!p)fa[las]=1;
        else
        {
            int q=ch[p][c];
            if(len[q]==len[p]+1)fa[las]=q;
            else
            {
                int qq=++tot;
                memcpy(ch[qq],ch[q],sizeof(ch[q]));
                fa[qq]=fa[q];
                len[qq]=len[p]+1;
                fa[q]=fa[las]=qq;
                for(;ch[p][c]==q;p=fa[p])ch[p][c]=qq;
            }
        }
        return las;
    }
    void bfs_build(Trie& trie)
    {
        queue<pii> q;
        q.em(pii(1,1));
        while(!q.empty())
        {
            auto pr=q.front(); q.pop();
            int pos=pr.first, u=pr.second;
            tb[pos]=u;
            for(int i=0;i<26;i++)
                if(trie[pos][i])
                {
                    int to=trie[pos][i], v=ch[u][i]?ch[u][i]:append(u,i,trie.end[to]);
                    q.em(pii(to,v));
                }
        }
    }
    int mov(int p, int c){return max(1,ch[p][c]);}
}exsam;
struct Graph
{
    int etot, head[maxn], to[maxe], next[maxe], w[maxe];
    void clear(int N)
    {
        for(int i=1;i<=N;i++)head[i]=0;
        etot=0;
    }
    void adde(int a, int b, int c=0){to[++etot]=b;w[etot]=c;next[etot]=head[a];head[a]=etot;}
    #define forp(_,__) for(auto p=__.head[_];p;p=__.next[p])
}G, G0;
struct Doubling_LCA
{
    int f[maxn][maxk+1], depth[maxn];
    void clear(int n){for(int i=1;i<=n;i++)depth[i]=0, cl(f[i]);}
    void dfs(Graph &G, int pos, int pre)
    {
        for(auto k=1;(1<<k)<=depth[pos];k++)f[pos][k]=f[f[pos][k-1]][k-1];
        for(auto p(G.head[pos]);p;p=G.next[p])
            if(G.to[p]!=pre)
            {
                f[G.to[p]][0]=pos;
                depth[G.to[p]]=depth[pos]+1;
                dfs(G,G.to[p],pos);
            }
    }
    void run(Graph &G, int root)
    {
        depth[root]=1;
        dfs(G,root,0);
    }
    int jp(int x, int b)
    {
        for(auto k=0;k<=maxk;k++)
            if(b&(1<<k))x=f[x][k];
        return x;
    }
}db;
void DP(Graph& G, vector<ll>& dp,  ll u)
{
    dp[u] = exsam.pref[u];
    forp(u,G)
    {
        ll v = G.to[p];
        DP(G,dp,v);
        dp[u] += dp[v];
    }
}
void dfs(Graph& G, Trie& T, ll ug, ll ut)
{
    tb[ug] = ut;
    forp(ug,G)
    {
        ll vg = G.to[p], alpha = s[vg] - 'A';
        ll vt = T.append(ut,alpha);
        dfs(G,T,vg,vt);
    }
}
int main()
{
    ll N=read(), Q=read(), i;
    vector<ll> fa(N+5), dp(2*N+5);
    trie.init();
    scanf("%s",s+2);
    fa[2]=1;
    rep(i,3,N+1)fa[i] = read()+1;
    rep(i,2,N+1)G0.adde(fa[i],i);
    dfs(G0,trie,1,1);
    exsam.init();
    exsam.bfs_build(trie);
    rep(i,2,exsam.tot)G.adde(exsam.fa[i],i);
    db.run(G,1);
    DP(G,dp,1);
    rep(i,1,Q)
    {
        ll X=tb[read()+1], L=read(), pos = exsam.tb[X], k;
        drep(k,maxk,0)
        {
            if(exsam.len[db.f[pos][k]]>=L)pos = db.f[pos][k];
        }
        printf("%lld\n",dp[pos]);
    }
    return 0;
}