HDU 2222 Keywords Search(AC自动机)题解

题意:给出模式串,问你主串出现了几种模式串。

思1路:ac自动机模板题。参考q学姐:av6295004

ac自动机的基本原理就是在trie上建Fail值,这样我每次失配就去找当前的最长后缀去匹配,一直到完全失配为止,这样我就找遍了所有的模式串。

代码:

#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include <iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 5e5 + 10;
const int maxM = 1e6 + 10;
const ull seed = 131;
const int INF = 0x3f3f3f3f;
const int MOD = 1e4 + 7;
struct Aho{
    struct state{
        int next[26];
        int fail, cnt;
    }node[maxn];
    int size;
    queue<int> q;

    void init(){
        for(int i = 0; i < maxn; i++){
            memset(node[i].next, 0, sizeof(node[i].next));
            node[i].fail = node[i].cnt = 0;
        }
        size =  1;
        while(!q.empty()) q.pop();
    }

    void insert(char *s){
        int len = strlen(s);
        int now = 0;
        for(int i = 0; i < len; i++){
            char c = s[i];
            if(node[now].next[c - 'a'] == 0)
                node[now].next[c - 'a'] = size++;
            now = node[now].next[c - 'a'];
        }
        node[now].cnt++;
    }

    void build(){
        node[0].fail = -1;
        q.push(0);

        while(!q.empty()){
            int u = q.front();
            q.pop();
            for(int i = 0; i < 26; i++){
                if(node[u].next[i]){
                    if(u == 0) node[node[u].next[i]].fail = 0;
                    else{
                        int v = node[u].fail;
                        while(v != -1){
                            if(node[v].next[i]){
                                node[node[u].next[i]].fail = node[v].next[i];
                                break;
                            }
                            v = node[v].fail;
                        }
                        if(v == -1) node[node[u].next[i]].fail = 0;
                    }
                    q.push(node[u].next[i]);
                }
            }
        }
    }

    int get(int u){ //匹配规则
        int ret = 0;
        while(u){
            ret += node[u].cnt;
            node[u].cnt = 0;
            u = node[u].fail;
        }
        return ret;
    }

    int match(char *s){
        int ret = 0, now = 0;
        int len = strlen(s);
        for(int i = 0; i < len; i++){
            char c = s[i];
            if(node[now].next[c - 'a'])
                now = node[now].next[c - 'a'];
            else{
                int p = node[now].fail;
                while(p != -1 && node[p].next[c - 'a'] == 0){
                    p = node[p].fail;
                }
                if(p == -1) now = 0;
                else now = node[p].next[c - 'a'];
            }
            if(node[now].cnt){
                ret += get(now);
            }
        }
        return ret;
    }
}ac;

char s[maxM];
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        int n;
        scanf("%d", &n);
        ac.init();
        for(int i = 0; i < n; i++){
            scanf("%s", s);
            ac.insert(s);
        }
        scanf("%s", s);
        ac.build();
        printf("%d
", ac.match(s));
    }
    return 0;
}
原文地址:https://www.cnblogs.com/KirinSB/p/11174904.html