学步园

题目连接：zoj 3430 Detect the Virus

题目大意：给定一个编码完的串，将每一个字符对应着表的数值转换成6位二进制，然后以8为一个数值，重新形成字符

串，判断给定询问串是否含有字符集中的串。

解题思路：主要是题意，逆编码部分注意，转换完了之后，可能有字符'\0'，所以不能用字符串的形式储存，要用int型

的数组。注意有相同串的可能。

#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;

const int maxn = 512 * 100;
const int sigma_size = 256;

struct Aho_Corasick {
    int sz, g[maxn][sigma_size];
    int tag[maxn], fail[maxn], last[maxn];
    bool vis[520];
    int jump[520];

    void init();
    int idx(char ch);
    void insert(int* str, int k);
    void getFail();
    int match(int* str);
    void put(int u);
}A;

inline int change(char ch) {
    if (ch >= 'A' && ch <= 'Z')
        return ch - 'A';
    if (ch >= 'a' && ch <= 'z')
        return ch - 'a' + 26;
    if (ch >= '0' && ch <= '9')
        return ch - '0' + 52;
    if (ch == '+')
        return 62;
    return 63;
}

const int maxl = 5005;

char w[maxl];
int s[maxl * 2];

void input(int* s) {

    scanf("%s", w);
    s[0]=0;  
    int n = strlen(w);
    while (n && w[n-1] == '=') n--;
    w[n] = 0;
    for(int i=0,len=0,x=0;w[i];++i){  
        len+=6,x=(x<<6)|change(w[i]);  
        if(len>=8){  
            s[++s[0]]=(x>>(len-8))&0xff;
            len-=8;   
        }  
    } 

    /*
       for(int i = 1; i <= s[0]; i++)
       printf(" %d", s[i]);
       printf("\n");
       */
}

int N, M;

int main () {
    while (scanf("%d", &N) == 1) {
        A.init();

        for (int i = 1; i <= N; i++) {
            input(s);
            A.insert(s, i);
        }
        A.getFail();

        scanf("%d", &M);
        for (int i = 1; i <= M; i++) {
            input(s);
            printf("%d\n", A.match(s));
        }
        printf("\n");
    }
    return 0;
}

void Aho_Corasick::init() {
    sz = 1;
    tag[0] = 0;
    memset(g[0], 0, sizeof(g[0]));
}

int Aho_Corasick::idx(char ch) {
    return ch;
}

void Aho_Corasick::put(int u) {
    int p = tag[u];
    while (p && vis[p] == 0) {
        vis[p] = 1;
        p = jump[p];
    }
    if (last[u])
        put(last[u]);
}

void Aho_Corasick::insert(int* str, int k) {
    int u = 0;

    for (int i = 1; i <= str[0]; i++) {
        int v = str[i];
        while (v >= sigma_size);
        if (g[u][v] == 0) {
            tag[sz] = 0;
            memset(g[sz], 0, sizeof(g[sz]));
            g[u][v] = sz++;
        }
        u = g[u][v];
    }

    jump[k] = tag[u];
    tag[u] = k;
}

int Aho_Corasick::match(int* str) {
    memset(vis, 0, sizeof(vis));
    int u = 0;

    for (int i = 1; i <= str[0]; i++) {
        int v = str[i];
        while (v >= sigma_size);
        while (u && g[u][v] == 0)
            u = fail[u];

        u = g[u][v];

        if (tag[u])
            put(u);
        else if (last[u])
            put(last[u]);
    }

    int ret = 0;
    for (int i = 1; i <= N; i++)
        if (vis[i])
            ret++;
    return ret;
}

void Aho_Corasick::getFail() {
    queue<int> que;

    for (int i  = 0; i < sigma_size; i++) {
        int u = g[0][i];
        if (u) {
            fail[u] = last[u] = 0;
            que.push(u);
        }
    }

    while (!que.empty()) {
        int r = que.front();
        que.pop();

        for (int i = 0; i < sigma_size; i++) {
            int u = g[r][i];

            if (u == 0) {
                g[r][i] = g[fail[r]][i];
                continue;
            }

            que.push(u);
            int v = fail[r];
            while (v && g[v][i] == 0)
                v = fail[v];

            fail[u] = g[v][i];
            last[u] = tag[fail[u]] ? fail[u] : last[fail[u]];
        }
    }
}