HDU2222 Keywords Search AC自动机模板


题面:

  • In the modern time, Search engine came into the life of everybody like Google, Baidu, etc.
    Wiskey also wants to bring this feature to his image retrieval system.
    Every image have a long description, when users type some keywords to find the image, the system will match the keywords with description of image and show the image which the most keywords be matched.
    To simplify the problem, giving you a description of image, and some keywords, you should tell me how many keywords will be match.

解题过程:


AC代码:

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
#define rep(i,l,p) for(int i=l;i<=p;i++)
#define fread() freopen("in.txt","r",stdin)
#define mp make_pair
#define Fi first
#define Se second
#define endl '\n'
typedef long long ll;
int T,n;
const int N = 1e4+10, L = 1e6+10;
char s[L];
int num;
struct Node
{
    int son[30];
    int fail,cnt;
}a[N*60];
queue<int> q;
void clear(int x){
    a[x].cnt = 0;
    a[x].fail = 0;
    memset(a[x].son,0,sizeof a[x].son);
}
void trie(char* c){
    int l = strlen(c);
    int x = 0;
    for(int i=0;i<l;i++){
        int t = c[i]-'a'+1;
        if(!a[x].son[t]){
            num++;
            clear(num);
            a[x].son[t] = num;
        }
        x = a[x].son[t];
    }
    a[x].cnt++;
}
void buildAC(){
    while(!q.empty()) q.pop();
    for(int i=1;i<=26;i++){
        if(a[0].son[i]) q.push(a[0].son[i]);
    }
    while(!q.empty()){
        int x = q.front();
        q.pop();
        int fail = a[x].fail;
        for(int i=1;i<=26;i++){
            int y = a[x].son[i];
            if(y){
                a[y].fail = a[fail].son[i];
                q.push(y);
            }else{
                a[x].son[i] = a[fail].son[i];
            }
        }
    }
    return;
}
int find(char *c){
    int l = strlen(c);
    int x = 0,ans = 0;
    for(int i=0;i<l;i++){
        int t = c[i]-'a'+1;
        while(x && !a[x].son[t]) x = a[x].fail;
        x = a[x].son[t];
        int p = x;
        while(p && a[p].cnt != -1){
            ans += a[p].cnt;
            a[p].cnt = -1;
            p = a[p].fail;
        }
    }
    return ans;
}
int main(int argc, char const *argv[])
{
    ios::sync_with_stdio(false);
    int T;
    cin >> T;
    while(T--){
        int n;
        cin >> n;
        num = 0;
        clear(0);
        for(int i=1;i<=n;i++){
            cin >> s;
            trie(s);
        }
        buildAC();
        cin >> s;
        cout << find(s) << endl;
    }
    return 0;
}

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转载自blog.csdn.net/Hagtaril/article/details/81564204