【2017 ACM/ICPC Asia Regional Qingdao Online 1003】hdu 6208 The Dominator of Strings

The Dominator of Strings

Time Limit: 3000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 753    Accepted Submission(s): 219

Problem Description

Here you have a set of strings. A dominator is a string of the set dominating all strings else. The stringS is dominated by T if S is a substring of T.

 

Input

The input contains several test cases and the first line provides the total number of cases.
For each test case, the first line contains an integer N indicating the size of the set.
Each of the following N lines describes a string of the set in lowercase.
The total length of strings in each case has the limit of 100000.
The limit is 30MB for the input file.

 

Output

For each test case, output a dominator if exist, or No if not.

 

Sample Input

310youbetterworsericherpoorersicknesshealthdeathfaithfulnessyoubemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulness5abccdeabcdeabcdebcde3aaaaaaaaabaaaac

 

Sample Output

youbemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulnessabcdeNo

 

AC自动机

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <map>#include <string>#include <cstring>#include <ctime>#include <queue>#define ms(a,b) memset(a,b,sizeof(a))using namespace std;typedef long long ll;const int N = 100010;const int M = 26;struct AC{    int ch[N][M], fail[N], val[N];    int sz;    void init()    {        sz = 1;        memset(ch[0], 0, sizeof(ch[0]));        memset(val, 0, sizeof(val));    }    int idx(char c)    {        return c - 'a';    }    void add(char *s)    {        int n, now = 0;        n = strlen(s);        for (int i = 0; i < n; i++)        {            int id = idx(s[i]);            if (ch[now][id] == 0)            {                memset(ch[sz], 0, sizeof(ch[sz]));                ch[now][id] = sz++;            }            now = ch[now][id];        }        val[now]++;    }    void Build()    {        queue<int>que;        fail[0] = 0;        for (int i = 0; i < M; i++)        {            int u = ch[0][i];            if (u)            {                fail[u] = 0;                que.push(u);            }        }        while (!que.empty())        {            int r = que.front();            que.pop();            for (int c = 0; c < M; c++)            {                int u = ch[r][c];                if (!u)                {                    continue;                }                que.push(u);                int v = fail[r];                while (v && !ch[v][c])                {                    v = fail[v];                }                fail[u] = ch[v][c];            }        }    }    int Query(char *s)    {        int n, now = 0, ans = 0;        n = strlen(s);        for (int i = 0; i < n; i++)        {            int id = idx(s[i]);            while (now && !ch[now][id])            {                now = fail[now];            }            now = ch[now][id];            int t = now;            while (t && val[t] != -1)            {                ans += val[t];                val[t] = -1;                t = fail[t];            }        }        return ans;    }} ac;char s[100010], a[100010];int main(){    ios::sync_with_stdio(false);    int T, n;    cin >> T;    while (T--)    {        cin >> n;        if (n == 0)        {            cout << "No" << endl;            continue;        }        int maxx = 0, hh = n, len;        ac.init();        while (hh--)        {            cin >> a;            len = strlen(a);            ac.add(a);            if (len > maxx)            {                maxx = len;                strcpy(s, a);                s[len] = '\0';            }        }        ac.Build();        if (ac.Query(s) == n)        {            cout << s << endl;        }        else        {            cout << "No" << endl;        }    }    return 0;}