题意:
有 n n n个字符串 s 1 , s 2 . . . s n s_1,s_2...s_n s1,s2...sn, q q q个操作:
1 Q k l r 1\ Q\ k\ l\ r 1 Q k l r,给出字符串 Q Q Q,求 s l , s l + 1 . . . s r s_l,s_{l+1}...s_r sl,sl+1...sr中和字符串 Q Q Q公共前缀长度大于等于 k k k的字符串有多少个。
2 a b 2\ a\ b 2 a b,交换 s a 和 s b s_a和s_b sa和sb
数据范围: n , q ≤ 2 e 5 , ∑ ∣ s ∣ ≤ 2 e 5 , ∑ ∣ Q ∣ ≤ 2 e 5 , k ≤ ∣ Q ∣ n,q\le 2e5,\ \sum|s|\le 2e5,\ \sum |Q|\le 2e5,k\le |Q| n,q≤2e5, ∑∣s∣≤2e5, ∑∣Q∣≤2e5,k≤∣Q∣
解题思路:
先对所有的 s s s建一颗字典树,每个 s s s都有一个对应的结点编号,设 Q Q Q的前 k k k的字母组成的字符串在字典树上对应的编号是 x x x,那么和 Q Q Q公共前缀大于等于 k k k的字符串对应的结点一定在以 x x x为根的子树里。那么把字典树按DFS序处理之后就变成了区间查询问题。问题就变成了单点修改,求 [ l , r ] [l,r] [l,r]内数值在 [ L , R ] [L,R] [L,R]之间的数有多少个。这样问题是不是就变得很熟悉了呢?(
这类问题可以用CDQ分治或者树套树解决,这里使用树状数组套线段树。
#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();
}