HDU 6208 The Dominator of Strings（AC自动机）

The Dominator of Strings

Time Limit: 3000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2830    Accepted Submission(s): 1010

Problem Description

Here you have a set of strings. A dominator is a string of the set dominating all strings else. The string S 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

3

10

you

better

worse

richer

poorer

sickness

health

death

faithfulness

youbemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulness

 

5

abc

cde

abcde

abcde

bcde

 

 

3

aaaaa

aaaab

aaaac

 

 

Sample Output

youbemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulness

abcde

No

 

 

 

题目链接：HDU 6208

检查你用的是不是假模板的好题……，经此一题，许多人发现自己学的是假AC自动机

代码：

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <string>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define fin(name) freopen(name,"r",stdin)
#define fout(name) freopen(name,"w",stdout)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 100010;
struct Trie
{
    int nxt[26];
    int cnt, fail;
    void init()
    {
        for (int i = 0; i < 26; ++i)
            nxt[i] = -1;
        cnt = fail = 0;
    }
} L[N];
char s[N];
int st[N], Len[N];
int tot;

namespace AC
{
    void init()
    {
        tot = 0;
        L[tot++].init();
    }
    void insert(char s[], int len)
    {
        int u = 0;
        for (int i = 0; i < len; ++i)
        {
            int v = s[i] - 'a';
            if (L[u].nxt[v] == -1)
            {
                L[tot].init();
                L[u].nxt[v] = tot++;
            }
            u = L[u].nxt[v];
        }
        ++L[u].cnt;
    }
    void build()
    {
        L[0].fail = 0;
        queue<int>Q;
        for (int i = 0; i < 26; ++i)
        {
            if (L[0].nxt[i] == -1)
                L[0].nxt[i] = 0;
            else
            {
                L[L[0].nxt[i]].fail = 0;
                Q.push(L[0].nxt[i]);
            }
        }
        while (!Q.empty())
        {
            int u = Q.front();
            Q.pop();
            int uf = L[u].fail;
            for (int i = 0; i < 26; ++i)
            {
                int v = L[u].nxt[i];
                if (v == -1)
                    L[u].nxt[i] = L[uf].nxt[i];
                else
                {
                    L[v].fail = L[uf].nxt[i];
                    Q.push(v);
                }
            }
        }
    }
    int query(char s[], int len)
    {
        int ret = 0;
        int u = 0;
        for (int i = 0; i < len; ++i)
        {
            int v = s[i] - 'a';
            u = L[u].nxt[v];
            while (u && L[u].nxt[v] == -1)
                u = L[u].fail;
            int t = u;
            while (t && L[t].cnt != -1)
            {
                ret += L[t].cnt;
                L[t].cnt = -1;
                t = L[t].fail;
            }
        }
        return ret;
    }
}
int main(void)
{
    int T, n, i;
    scanf("%d", &T);
    while (T--)
    {
        AC::init();
        scanf("%d", &n);
        int sum = 0;
        int Maxlen = 0;
        int ID = 0;
        for (i = 0; i < n; ++i)
        {
            scanf("%s", s + sum);
            Len[i] = strlen(s + sum);
            AC::insert(s + sum, Len[i]);
            st[i] = sum;
            sum += Len[i];
            if (Len[i] > Maxlen)
            {
                ID = i;
                Maxlen = Len[i];
            }
        }
        AC::build();
        if (AC::query(s + st[ID], Len[ID]) == n)
        {
            int ed = st[ID] + Len[ID];
            for (i = st[ID]; i < ed; ++i)
                printf("%c", s[i]);
            puts("");
        }
        else
            puts("No");
    }
    return 0;
}