Nobita's New Filesystem （bitset）

因为题目要求的是必须要全部拥有，那么天生就有一种与的性质，然后就会想出一种暴力做法：

对于每一个标签，用一个bitset存储哪些文件有它

然后查询就是把所有的标签的bitset取个and就是答案集合

然后就过了

#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 ;
}

/*
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？(正難則反)
*/