AC自动机板子

例题：HDU - 2222

给(n)个字符串，一个模式串。然后输出匹配次数。

代码

#include <bits/stdc++.h>

#define FOPI freopen("in.txt", "r", stdin)
#define FOPO freopen("out.txt", "w", stdout)
#define FOR(i,x,y) for (int i = x; i <= y; i++)
#define ROF(i,x,y) for (int i = x; i >= y; i--)

using namespace std;
typedef long long LL;

struct Trie
{
    int next[500010][26], fail[500010], end[500010];
    int root, L;
    int newnode()
    {
        FOR(i, 0, 26-1) next[L][i] = -1;
        end[L++] = 0;
        return L-1;
    }
    void init() { L = 0; root = newnode(); }
    void insert(char buf[])
    {
        int len = strlen(buf), now = root;
        FOR(i, 0, len-1)
        {
            if (next[now][buf[i] - 'a'] == -1)
                next[now][buf[i] - 'a'] = newnode();
            now = next[now][buf[i] - 'a'];
        }
        end[now]++;
    }
    void build()
    {
        queue<int> Q;
        fail[root] = root;
        FOR(i, 0, 26-1)
            if (next[root][i] == -1) next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            FOR(i, 0, 26-1)
                if (next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }

    int query(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        int res = 0;
        FOR(i, 0, len-1)
        {
            now = next[now][buf[i]-'a'];
            int temp = now;
            while(temp != root)
            {
                res += end[temp];
                end[temp] = 0;
                temp = fail[temp];
            }
        }
        return res;
    }

//    void debug()
//    {
//        FOR(i, 0, L-1)
//        {
//            printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], end[i]);
//            FOR(j, 0, 26-1) printf("%2d", next[i][j]);
//            printf("]
");
//        }
//    }
};


char buf[1000010];
Trie ac;
int t, n;

int main()
{
    scanf("%d", &t);
    FOR(ca, 1, t)
    {
        scanf("%d", &n);
        ac.init();
        FOR(i, 0, n-1)
        {
            scanf("%s", buf);
            ac.insert(buf);
        }
        ac.build();
        scanf("%s", buf);
        printf("%d
", ac.query(buf));
    }
}