2019 ICPC Nanjing Regional F. Paper Grading (DFS sequential partition of dictionary tree + tree nesting tree)

Title:

Have $n$ strings$s_1,s_2...s_n$，$q$ operations:
$1Qklr$ , gives the string$Q$，求$s_l,s_{l+1}...s_r$Neutralization string $Q$ common prefix length is greater than or equal tokkHow many strings are there in$k$ ?
$2ab$ , exchange$s_{a}And{s}_b$
Data range: $n,q\le 2e5,\sum |s|\le 2e5,\sum |Q|\le 2e5,k\le |Q|$

Problem-solving ideas:

First for all $s$ build a dictionary tree, each$s$ has a corresponding node number, set$Q$ 's former$The$ corresponding number on the dictionary tree for a string composed of letters of$k$ is$x$ , then and$Q$ common prefix is ​​greater than or equal tokkThe node corresponding to the string of$k$ must be in$x$ is the root of the subtree. After processing the dictionary tree in DFS order, it becomes an interval query problem. The problem becomes a single point of modification, find$[l,r]$ inner value is in$[L,$How many numbers are there between$R$$]$ ? Does this become familiar to the problem? (
This kind of problem can be solved with CDQ divide and conquer or tree nest tree. Here, a tree array nested line segment tree is used.

#include<bits/stdc++.h>
#define ll long long
#define lowbit(x) ((x)&(-(x)))
#define mid ((l+r)>>1)
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
using namespace std;
const int maxn = 2e5 + 50;
int ch[maxn][26], sz[maxn], idx = 0;
int T[maxn], val[maxn*200], lc[maxn*200], rc[maxn*200], tot;
int n, q;
void update(int &rt, int l, int r, int pos, int x){
    
    
    if(!rt) rt = ++tot;
    val[rt]+=x;
    if(l == r) return;
    if(pos <= mid) update(lc[rt], l, mid, pos, x);
    else update(rc[rt], mid+1, r, pos, x);
    return;
}
int qry(int rt, int l, int r, int L, int R){
    
    
    if(!rt) return 0;
    if(L <= l && r <= R) return val[rt];
    int res = 0;
    if(L <= mid) res += qry(lc[rt], l, mid, L, R);
    if(R > mid) res += qry(rc[rt], mid+1, r, L, R);
    return res;
}
void add(int i, int pos, int x){
    
    
    while(i <= n){
    
    
        update(T[i], 1, idx, pos, x);
        i += lowbit(i);
    }return;
}
int sum(int i, int L, int R){
    
    
    int res = 0;
    while(i){
    
    
        res += qry(T[i], 1, idx, L, R);
        i -= lowbit(i);
    }return res;
}
int root, cnt;
int add(char *s){
    
    
    int p = root;
    while(*s){
    
    
        int x = *s-'a';
        if(!ch[p][x]) ch[p][x] = ++cnt;
        p = ch[p][x];
        s++;
    }return p;
}
int dfn[maxn], id[maxn];
void dfs(int u){
    
    
    dfn[++idx] = u; id[u] = idx;
    sz[u] = 1;
    for(int i = 0; i < 26; ++i){
    
    
        if(ch[u][i]) dfs(ch[u][i]), sz[u] += sz[ch[u][i]];
    }return;
}

char s[maxn];
int to[maxn];
void init(){
    
    
    scanf("%d%d", &n, &q);
    cnt = 0; root = ++cnt;
    for(int i = 1; i <= n; ++i){
    
    
        scanf("%s", s); to[i] = add(s);
        //cout<<"i:"<<i<<" t:"<<to[i]<<endl;
    }
    dfs(root);
    //cout<<"idx:"<<idx<<endl;
    for(int i = 1; i <= n; ++i){
    
    
        add(i, id[to[i]], 1);
    }
}
int get_to(int k){
    
    
    int p = root;
    for(int i = 0; i < k; ++i){
    
    
        int x = s[i]-'a';
        if(!ch[p][x]) return -1;
        p = ch[p][x];
    }return p;
}
void sol(){
    
    
    while(q--){
    
    
        int op; scanf("%d", &op);
        if(op == 1){
    
    
            int u, v; scanf("%d%d", &u, &v);
            add(u, id[to[u]], -1);
            add(u, id[to[v]], 1);
            add(v, id[to[v]], -1);
            add(v, id[to[u]], 1);
            swap(to[u], to[v]);
        }else{
    
    
            int k, l, r, L, R;
            scanf("%s", s);
            scanf("%d%d%d", &k, &l, &r);
            int u = get_to(k);
            if(u == -1){
    
    
                printf("0\n"); continue;
            }
            L = id[u], R = id[u]+sz[u]-1;
            //cout<<"L:"<<L<<" R:"<<R<<endl;
            int ans = sum(r, L, R)-sum(l-1, L, R);
            printf("%d\n", ans);
        }
    }
}
int main()
{
    
    
    init();
    sol();
}