题解 | #子串查询#

贪心，和哈希没关系

#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
//如果在不支持 avx2 的平台上将 avx2 换成 avx 或 SSE 之一
#include<bits/stdc++.h>
using namespace std;
#define x first
#define y second
typedef pair<int,int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef vector<string> VS;
typedef vector<int> VI;
typedef vector<vector<int>> VVI;
vector<int> vx;
inline void divide() {sort(vx.begin(),vx.end());vx.erase(unique(vx.begin(),vx.end()),vx.end());}
inline int mp(int x) {return upper_bound(vx.begin(),vx.end(),x)-vx.begin();}
inline int log_2(int x) {return 31-__builtin_clz(x);}
inline int popcount(int x) {return __builtin_popcount(x);}
inline int lowbit(int x) {return x&-x;}
inline ll Lsqrt(ll x) { ll L = 1,R = 2e9;while(L + 1 < R){ll M = (L+R)/2;if(M*M <= x) L = M;else R = M;}return L;}
set<int> pos[26];
void solve()
{
    int n,q;
    cin>>n>>q;
    VS g(q+1);
    for(int i = 0; i <= q; ++i) cin>>g[i];
    for(int i = 0; i < n; ++i) pos[g[0][i] - 'a'].insert(i);
    for(int i = 1; i <= q; ++i)
    {
        int t = g[i].size();
        int last = -1;
        bool ok = true;
        for(int j = 0;j < t; ++j)
        {
            auto it = pos[g[i][j]-'a'].upper_bound(last);
            if(it == pos[g[i][j]-'a'].end())
                ok = false;
            else last = *it;
        }
        cout<<(ok?("YES\n"):("NO\n"));
    }
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int T = 1;
	//cin>>T;
	while(T--)
	{
		solve();
	}
}