[BZOJ 2894]世界线

传送门

\(\color{green}{solution}\)

在开这道题之前建议先看看3756:pty的字符串,然后你会发现这题就很zz了.
当然，作为一名合格的博主,我还是应该写点什么的

首先,我们要清楚广义后缀自动机不是什么高大上的东西,就是普通的后缀自动机建完一个串后
把\(last\)的值赋为\(root\)建下一个串...(然而我搞了一星期才搞懂).
然后这是一棵数,在树上建后缀自动机的话就把当前节点作为\(last\),然后把子节点正常插入即可
于是这道题就变成了[TJOI2015]弦论
剩下的就是一堆的细节问题,比如他让你求的是本质不同的第\(K\)大,然后空串要算一个单独的串.
最为重点的细节是不要用\n,换成puts("").至于为什么我也不太清楚,但是看了我的
提交记录你就明白了

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

const int maxn = 1000010;

typedef long long ll;

int ch[maxn][26], fa[maxn], dis[maxn], cnt = 1, lst = 1;

int insert(int p, int val) {
    int x = ch[p][val];
    if( !x) {
        int np = ++ cnt; dis[np] = dis[p] + 1; lst = np;
        for ( ; p && !ch[p][val]; p = fa[p]) ch[p][val] = np;
        if( !p) fa[np] = 1;
        else {
            int q = ch[p][val];
            if( dis[q] == dis[p] + 1) fa[np] = q;
            else {
                int nq = ++ cnt; dis[nq] = dis[p] + 1;
                memcpy(ch[nq], ch[q], sizeof(ch[q]));
                fa[nq] = fa[q]; fa[q] = fa[np] = nq;
                for ( ; ch[p][val] == q; p = fa[p]) ch[p][val] = nq;
            }
        }
    }
    else {
        int q = ch[p][val];
        if( dis[q] == dis[p] + 1) lst = q;
        else {
            int nq = ++ cnt; dis[nq] = dis[p] + 1;
            lst = nq;
            memcpy(ch[nq], ch[q], sizeof(ch[q]));
            fa[nq] = fa[q]; fa[q] = nq;
            for ( ; ch[p][val] == q; p = fa[p]) ch[p][val] = nq;
        }
    }
    return lst;
}

int c[maxn], g[maxn];

vector<int> edge[maxn];

char s[maxn];

inline void dfs(int u, int pre, int p) {
    p = insert(p, s[u] - 'a');
    for ( int i = 0; i < edge[u].size(); ++ i) if( edge[u][i] ^ pre) {
        dfs(edge[u][i], u, p);
    }
}

int n, m;

ll sz[maxn];

inline void out(int u, ll kth) {
    if( !( --kth)) return;
    for ( register int i = 0; i <= 25; ++ i) {
        if( kth > sz[ch[u][i]]) kth -= sz[ch[u][i]];
        else {
            putchar('a'+i); out(ch[u][i], kth); return;
        }
    }
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("1.in", "r", stdin);
    freopen("a.out", "w", stdout);
#endif
    scanf("%d%d%s", &n, &m, s+1);
    for ( register int i = 2, a, b; i <= n; ++ i) {
        scanf("%d%d", &a, &b);
        edge[a].push_back(b); edge[b].push_back(a);
    }
    dfs(1, 0, 1);
    for ( register int i = 1; i <= cnt; ++ i) c[dis[i]] ++;
    for ( register int i = 1; i <= cnt; ++ i) c[i] += c[i-1];
    for ( register int i = 1; i <= cnt; ++ i) g[c[dis[i]] --] = i;
    for ( register int i = cnt; i; -- i) {
        sz[g[i]] = 1;
        for ( register int k = 0; k <= 25; ++ k) sz[g[i]] += sz[ch[g[i]][k]];
    }
    printf("%lld", sz[1]); 
    puts("");
    for ( register int i = 1; i <= m; ++ i) {
        ll K; scanf("%lld", &K);
        if( K > sz[1]) { puts("-1"); continue;}
        out(1, K); puts("");
    }
    return 0;
}

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