CF701F String set queries (分块思想+暴力)

很容易想到AC自动机,但是却发现不怎么支持删除

完蛋,怎么办?

思考如何优化暴力

有两种暴力:一种是kmp,一种是trie

trie时间复杂度优秀,但空间不行;

kmp时间不行

那么我们可以互补一下

对于长度小于 \(sqrt\) 的,我们加入 \(trie\) 中,否则暴力 \(kmp\),这样能够维持时间复杂度,因为总长不大

分块思想真妙!

#include <map>
#include <set>
#include <ctime>
#include <queue>
#include <stack>
#include <cmath>
#include <vector>
#include <bitset>
#include <cstdio>
#include <cctype>
#include <string>
#include <numeric>
#include <cstring>
#include <cassert>
#include <climits>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std ;
//#define int long long
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
#define per(i, a, b) for (int i = (a); i >= (b); i--)
#define loop(s, v, it) for (s::iterator it = v.begin(); it != v.end(); it++)
#define cont(i, x) for (int i = head[x]; i; i = e[i].nxt)
#define clr(a) memset(a, 0, sizeof(a))
#define ass(a, sum) memset(a, sum, sizeof(a))
#define lowbit(x) (x & -x)
#define all(x) x.begin(), x.end()
#define ub upper_bound
#define lb lower_bound
#define pq priority_queue
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define iv inline void
#define enter cout << endl
#define siz(x) ((int)x.size())
#define file(x) freopen(#x".in", "r", stdin),freopen(#x".out", "w", stdout)
typedef long long ll ;
typedef unsigned long long ull ;
typedef pair <int, int> pii ;
typedef vector <int> vi ;
typedef vector <pii> vii ;
typedef queue <int> qi ;
typedef queue <pii> qii ;
typedef set <int> si ;
typedef map <int, int> mii ;
typedef map <string, int> msi ;
const int N = 300010 ;
const int Sqrt = 500 ;
const int M = N / Sqrt + 10 ;
const int INF = 0x3f3f3f3f ;
const int iinf = 1 << 30 ;
const ll linf = 2e18 ;
const int MOD = 1000000007 ;
const double eps = 1e-7 ;
void print(int x) { cout << x << endl ; exit(0) ; }
void PRINT(string x) { cout << x << endl ; exit(0) ; }
void douout(double x){ printf("%lf\n", x + 0.0000000001) ; }
template <class T> void chmin(T &a, T b) { if (a > b) a = b ; }
template <class T> void chmax(T &a, T b) { if (a < b) a = b ; }
void upd(int &a, int b) { (a += b) %= MOD ; }
void mul(int &a, int b) { a = 1ll * a * b % MOD ; }

char s[N] ;
string st[M] ;
int cost[M] ;
int trie[N][29], value[N], nxt[N] ;
int Q, op, tot = 1, cnt, len ;

signed main(){
//  freopen("test.in", "r", stdin) ;
//  freopen("test.out", "w", stdout) ;
    scanf("%d", &Q) ;
    while (Q--) {
        scanf("%d %s", &op, s) ;
        len = strlen(s) ;
        if (op < 3) {
            if (len <= Sqrt) {
                int now = 1 ;
                rep(i, 0, len - 1) {
                    int c = s[i] - 'a' ;
                    if (!trie[now][c]) trie[now][c] = ++tot ;
                    now = trie[now][c] ;
                }
                value[now] += 3 - 2 * op ;
            } else {
                st[cnt] = "" ;
                rep(j, 0, len - 1) st[cnt] += s[j] ;
                st[cnt] += '{' ;
                cost[cnt] = 3 - 2 * op ;
                cnt++ ;
            }
        } else {
            ll ans = 0 ;
            rep(j, 0, len - 1) {
                int now = 1 ;
                rep(k, j, len - 1) {
                    int c = s[k] - 'a' ;
                    if (!trie[now][c]) break ;
                    now = trie[now][c] ;
                    ans += value[now] ;
                }
            }
        //  cout << ans << endl ;
            rep(id, 0, cnt - 1) {
                int Len = siz(st[id]) - 1 ; // 有一个结尾符
                if (Len > len) continue ;
                nxt[1] = 0 ;
                for (int i = 2, j = 0; i <= Len; i++) {
                    while (j && st[id][i - 1] != st[id][j]) j = nxt[j] ;
                    if (st[id][i - 1] == st[id][j]) j++ ;
                    nxt[i] = j ;
                }
                for (int i = 1, j = 0; i <= len; i++) {
                    while (j && s[i - 1] != st[id][j]) j = nxt[j] ;
                    if (s[i - 1] == st[id][j]) j++ ;
                    if (j == Len) ans += cost[id] ;
                }
            }
            cout << ans << endl ;
        }
    }
    return 0 ;
}

/*
写代码时请注意:
    1.ll?数组大小,边界?数据范围?
    2.精度?
    3.特判?
    4.至少做一些
思考提醒:
    1.最大值最小->二分?
    2.可以贪心么?不行dp可以么
    3.可以优化么
    4.维护区间用什么数据结构?
    5.统计方案是用dp?模了么?
    6.逆向思维?
*/


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转载自www.cnblogs.com/harryhqg/p/10548925.html
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