【题目链接】:http://hihocoder.com/problemset/problem/1036?sid=1092555

【题意】

【题解】



AC自动机的模板题;

在求有没有子串的时候;

注意要遍历所有后缀相同的情况;不然会漏解;

然后之前找过的就不要再找一遍了(即从那个状态找不能找到某个单词的终点);



【Number Of WA】



1



【完整代码】

#include <bits/stdc++.h>

using namespace std;

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define LL long long

#define rep1(i,a,b) for (int i = a;i <= b;i++)

#define rep2(i,a,b) for (int i = a;i >= b;i--)

#define mp make_pair

#define pb push_back

#define fi first

#define se second

#define ms(x,y) memset(x,y,sizeof x)

typedef pair<int,int> pii;

typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};

const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};

const double pi = acos(-1.0);

const int N = 110;

const int MAX = 1e6+100;

int n,a[MAX][27],f[MAX],cnt[MAX],tot = 1,flag[MAX];

queue <int> dl;

string s;

int main()

{

    //freopen("F:\\rush.txt","r",stdin);

    ios::sync_with_stdio(false),cin.tie(0);//scanf,puts,printf not use

    rep1(i,1,26) a[0][i] = 1;

    cin >> n;

    rep1(i,1,n)

    {

        cin >> s;

        int len = s.size();

        int now = 1;

        rep1(j,0,len-1)

        {

            int t = s[j]-'a'+1;

            if (a[now][t]==0) a[now][t]=++tot;

            now = a[now][t];

        }

        cnt[now]++;

    }

    f[1] = 0;

    dl.push(1);

    while (!dl.empty())

    {

        int x = dl.front();

        dl.pop();

        rep1(i,1,26)

        {

            if (a[x][i]==0) continue;

            int now = f[x];

            while (!a[now][i]) now = f[now];

            f[a[x][i]] = a[now][i];

            dl.push(a[x][i]);

        }

    }

    cin >> s;

    int len = s.size();

    int now = 1;

    rep1(i,0,len-1)

    {

        int t = s[i]-'a'+1;

        while (!a[now][t]) now = f[now];

        now = a[now][t];

        if (!flag[now])

        {

            for (int j = now;j;j=f[j])

            {

                if (flag[j]) break;

                flag[j] = true;

                if (cnt[j])

                    return cout <<"YES"<<endl,0;

            }

        }

    }

    cout <<"NO"<<endl;

    return 0;

}