HDU 3065

    又一AC自动机模板题...

#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 long

using namespace std;

struct AcAutoMachine {
#define N 50005     // 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, cunt;
    int queu[N], reco[1005], head, tail;
    
    int code(const char &c) {   // 将字符哈希成结点标号
        if (c >= 'A' && c <= 'Z')
            return c - 'A';
        else
            return -1;
    }
    void clear(int x) {
        memset(next[x], -1, sizeof(next[x]));
        fail[x] = -1; data[x] = 0;
    }
    void init() {
        clear(0);
        size = cunt = 1;
        memset(reco, 0, sizeof(reco));
    }
    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] = cunt++;
    }
    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];
                }
            }
        }
    }
    void query(char *str) {
        int p = 0, ret = 0, idx, pp;
        while (*str) {
            idx = code(*str);
            if (idx == -1) {
                p = 0;
            } else {
                p = next[p][idx];
                for (pp = p; pp != -1; pp = fail[pp]) {
                    ret += data[pp];
                    if (data[pp]) reco[data[pp]]++;
                }
            }
            ++str;
        }
    }
} A;

char name[1005][51];
char str[2000005];

int main() {
    int n, i;
    while (scanf("%d", &n) != EOF) {
        A.init();
        for (i = 1; i <= n; i++) {
            scanf("%s", name[i]);
            A.insert(name[i]);
        }
        A.buildac();
        getchar();
        gets(str);
        A.query(str);
        for (i = 1; i <= n; i++)
            if (A.reco[i]) printf("%s: %d\n", name[i], A.reco[i]);
    }
    return 0;
}