LA 4670 Dominating Patterns (AC自动机)

Posted 2020-08-17 dwtfukgv

题意：给定一个一篇文章，然后下面有一些单词，问这些单词在这文章中出现过几次。

析：这是一个AC自动机的裸板，最后在匹配完之后再统计数目就好。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 5;
const LL mod = 2147493647;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){  return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
const int sigma = 26;
const int maxnode = 70 * 150 + 5;

struct Aho{
    int cnt[155];
    int ch[maxnode][sigma];
    int f[maxnode];
    int val[maxnode];
    int last[maxnode];
    int sz;
    void init(){
        sz = 1;
        memset(ch[0], 0, sizeof ch[0]);
        memset(cnt, 0, sizeof cnt);
    }

    int idx(char ch){  return ch - ‘a‘; }

    void print(int j){
        if(j){
            ++cnt[val[j]];
            print(last[j]);
        }
    }
    void insert(char *s, int v){
        int u = 0;
        while(*s){
            int c = idx(*s);
            if(!ch[u][c]){
                memset(ch[sz], 0, sizeof ch[sz]);
                val[sz] = 0;
                ch[u][c] = sz++;
            }
            u = ch[u][c];  ++s;
        }
        val[u] = v;

    }

    void find(char *s){
        int j = 0;
        while(*s){
            int c = idx(*s);
            while(j && !ch[j][c]) j = f[j];
            j = ch[j][c];
            if(val[j]) print(j);
            else if(last[j])  print(last[j]);
            ++s;
        }
    }

    void getFail(){
        queue<int> q;
        f[0] = 0;
        for(int c = 0; c < sigma; ++c){
            int u = ch[0][c];
            if(u){ f[u] = 0;  q.push(u);  last[u] = 0; }
        }

        while(!q.empty()){
            int r = q.front();  q.pop();
            for(int c = 0;c < sigma; ++c){
                int u = ch[r][c];
                if(!u)  continue;
                q.push(u);
                int v = f[r];
                while(v && !ch[v][c]) v = f[v];
                f[u] = ch[v][c];
                last[u] = val[f[u]] ? f[u] : last[f[u]];
            }
        }
    }
};

Aho aho;
char s[155][75];
char t[1000005];
map<string, int> mp;

int main(){
    while(scanf("%d", &n) == 1 && n){
        aho.init();
        mp.clear();
        for(int i = 1; i <= n; ++i){
            scanf("%s", s[i]);
            aho.insert(s[i], i);
            mp[s[i]] = i;
        }
        scanf("%s", t);
        aho.getFail();
        aho.find(t);
        int mmax = -1;
        for(int i = 1; i <= n; ++i)  mmax = Max(mmax, aho.cnt[i]);
        printf("%d\n", mmax);
        for(int i = 1; i <= n; ++i)
            if(aho.cnt[mp[s[i]]] == mmax)  printf("%s\n", s[i]);
    }
    return 0;
}