Nobita's New Filesystem （bitset）

由於題目要求的是必需要所有擁有，那麼天生就有一種與的性質，而後就會想出一種暴力作法：

對於每個標籤，用一個bitset存儲哪些文件有它

而後查詢就是把全部的標籤的bitset取個and就是答案集合

而後就過了

#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 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 = 100010 ;
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 ; }
template <class T> void upd(T &a, T b) { (a += b) %= MOD ; }
template <class T> void mul(T &a, T b) { a = 1ll * a * b % MOD ; }

string sta ;
map <string, bitset <1024> > mp ;
map <string, bitset <1024> >::iterator it ;
bitset <1024> mask ;
char s[N] ;
ll sz[N] ;
int n, m ;

signed main() {
    while (scanf("%d", &n) != EOF) {
        mp.clear() ;
        rep(i, 1, n) {
            scanf("%s%lld", s, &sz[i - 1]) ;
            sta = "" ; int len = strlen(s) ;
            rep(j, 0, len - 1) {
                if (s[j] == '[') {
                    sta = "" ;
                }
                else if (s[j] == ']') {
        //          cout << sta << endl ;
                    mp[sta].set(i - 1) ;
                } else {
                    sta += s[j] ;
                }
            }
        }
        scanf("%d", &m) ;
        rep(i, 1, m) {
            scanf("%s", s) ; int len = strlen(s) ;
            mask.reset() ;
            rep(j, 0, n - 1) mask.set(j) ; // full
            sta = "" ;
            rep(j, 0, len - 1) {
                if (s[j] == '[') {
                    sta = "" ;
                }
                else if (s[j] == ']') {
                    it = mp.find(sta) ;
         //         cout << " * " << sta << endl ;
                    if (it == mp.end()) {
                        mask.reset() ;
                        break ;
                    } else {
                        mask &= it->se ;
                    }
                } else {
                    sta += s[j] ;
                }
            }
            ll ans = 0 ;
            rep(j, 0, n - 1) if (mask[j]) ans += sz[j] ;
            printf("%lld\n", ans) ;
        }
    }

    return 0 ;
}

/*
寫代碼時請注意：
    1.ll？數組大小，邊界？數據範圍？
    2.精度？
    3.特判？
    4.至少作一些
思考提醒：
    1.最大值最小->二分？
    2.能夠貪心麼？不行dp能夠麼
    3.能夠優化麼
    4.維護區間用什麼數據結構？
    5.統計方案是用dp？模了麼？
    6.逆向思惟？(正難則反)
*/