LA 4670 出现次数最多的子串 （AC自动机模板题）

Posted 2020-08-04 HARD_UNDERSTAND

Dominating Patterns

Time Limit:3000MS

Memory Limit:Unknown

64bit IO Format:%lld & %llu

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Description

The archaeologists are going to decipher a very mysterious ``language". Now, they know many language patterns; each pattern can be treated as a string on English letters (only lower case). As a sub string, these patterns may appear more than one times in a large text string (also only lower case English letters).

What matters most is that which patterns are the dominating patterns. Dominating pattern is the pattern whose appearing times is not less than other patterns.

It is your job to find the dominating pattern(s) and their appearing times.

Input

The entire input contains multi cases. The first line of each case is an integer, which is the number of patternsN, 1N150. Each of the following N lines contains one pattern, whose length is in range [1, 70]. The rest of the case is one line contains a large string as the text to lookup, whose length is up to10⁶.

At the end of the input file, number `0‘ indicates the end of input file.

Output

For each of the input cases, output the appearing times of the dominating pattern(s). If there are more than one dominating pattern, output them in separate lines; and keep their input order to the output.

Sample Input

2 
aba 
bab 
ababababac 
6 
beta 
alpha 
haha 
delta 
dede 
tata 
dedeltalphahahahototatalpha 
0

Sample Output

4 
aba 
2 
alpha 
haha

题意：
有n个小写字母组成的字符床和一个文本串   你的任务是找出哪些字符串在文本中出现的次数最多   例如  aba 在ababa中出现2次  但是bab只出现了一次
输入n  之后n个字符串   长度为1-70   n小于等于150  之后一个文本串  长度最长为10的6次方    
输出出现最多的次数 以及出现最多的字符串为什么  如果存在多个  按输入顺序排列

分析：
AC自动机模板题

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
#define rep(i,s,t) for(int i=(s);i<(t);i++)
const int INF = 1e9 + 9;
const int N = 150 * 70 + 9;
const int M = 26;
char str[155][77];
struct Trie {

    int nex[N][M], fail[N], val[N];
    int rt, L;
    int newnode() {
        memset (nex[L], -1, sizeof (nex[L]) );
        val[L++] = -1;
        return L - 1;
    }
    void init() {
        L = 0;
        rt = newnode();
    }
    int idx (char c) {
        return c - ‘a‘;
    }
    void insert (char s[], int id) {
        int len = strlen (s);
        int now = rt;
        for (int i = 0; i < len; i++) {
            if (nex[now][idx (s[i])] == -1)
                nex[now][idx (s[i])] = newnode();
            now = nex[now][idx (s[i])];
        }
        val[now] = id;
    }
    void getFail() {
        queue<int>q;
        fail[rt] = rt;
        for (int i = 0; i < M; i++)
            if (nex[rt][i] == -1) nex[rt][i] = rt;
            else {
                fail[nex[rt][i]] = rt;
                q.push (nex[rt][i]);
            }
        while (!q.empty() ) {
            int now = q.front();
            q.pop();
            for (int i = 0; i < M; i++)
                if (nex[now][i] == -1)
                    nex[now][i] = nex[fail[now]][i];
                else {
                    fail[nex[now][i]] = nex[fail[now]][i];
                    q.push (nex[now][i]);
                }
        }
    }
    int num[155];
    void quert (char s[], int n) {
        memset (num, 0, sizeof (num) );
        int len = strlen (s);
        int now = rt;
        for (int i = 0; i < len; i++) {
            now = nex[now][idx(s[i])];
            int tmp = now;
            while (tmp != rt) {
                if (val[tmp] != -1)
                    num[val[tmp]]++;
                tmp = fail[tmp];
            }
        }
        int maxn=0;
        for(int i=0;i<n;i++)maxn=max(num[i],maxn);
        printf("%d\n",maxn);
        for (int i = 0; i < n; i++)
            if (num[i]==maxn) printf ("%s\n", str[i]);
    }
};
char buf[1000009];
Trie ac;
int main() {
   // freopen("f.txt","r",stdin);
    int n;
    while (~scanf ("%d", &n)&&n ) {
        ac.init();
        for (int i = 0; i < n; i++) {
            scanf ("%s", str[i]);
            ac.insert (str[i], i);
        }
        ac.getFail();
        scanf ("%s", buf);
        ac.quert (buf, n);
    }
    return 0;
}