HDU 2222

来源：互联网发布：指画用什么软件编辑：程序博客网时间：2024/05/17 07:29
AC自动机模板...
#include <stdio.h>#include <stdlib.h>#include <math.h>#include <string.h>#include <string>#include <vector>#include <map>#include <set>#include <queue>#include <algorithm>#include <iostream>#include <sstream>#define ll long longusing namespace std;struct AcAutoMachine {#define N 500005    // 字典的节点数，比最大数据略大即可#define M 26        // 字典的字母数，此为精确值    int next[N][M], fail[N], data[N], size;    int queu[N], head, tail;        int code(const char &c) {   // 将字符哈希成结点标号        return c - 'a';    }    void clear(int x) {        memset(next[x], -1, sizeof(next[x]));        fail[x] = -1; data[x] = 0;    }    void init() {        clear(0);        size = 1;    }    void insert(char *str) {        int p = 0, idx;        while (*str) {            idx = code(*str);            if (next[p][idx] == -1) {                clear(size);                next[p][idx] = size++;            }            p = next[p][idx];            str++;        }        data[p]++;    }    void buildac() {        int p, pp, fp, fpp;        head = tail = 0;        queu[tail++] = 0;        while (head < tail) {            p = queu[head++];            for (int i = 0; i < M; i++) {                if ((pp = next[p][i]) != -1) {                    for (fp = fail[p]; fp != -1; fp = fail[fp]) {                        if ((fpp = next[fp][i]) != -1) {                            fail[pp] = fpp;                            // 将fail链上后继结点的权值加到前驱结点中，                            // 这样计算权值时就不需要遍历fail链，看情况是否需要。                            // data[pp] += data[fpp];                            break;                        }                    }                    if (fp == -1) fail[pp] = 0;                    queu[tail++] = pp;                } else {    // 重构next，看题目是否需要                    if (p == 0) next[p][i] = 0;                    else next[p][i] = next[fail[p]][i];                }            }        }    }    int query(char *str) {        int p = 0, ret = 0, idx, pp;        while (*str) {            idx = code(*str);            p = next[p][idx];            for (pp = p; pp != -1 && data[pp] != -1; pp = fail[pp]) {                ret += data[pp];                data[pp] = -1;            }            ++str;        }        return ret;    }} A;char str[1000005];int main() {    int n, i, t;    for (scanf("%d", &t); t--; ) {        scanf("%d", &n);        A.init();        for (i = 0; i < n; i++) {            scanf("%s", str);            A.insert(str);        }        A.buildac();        scanf("%s", str);        printf("%d\n", A.query(str));    }    return 0;}