zju3430

题意：给出多个加密的模式串，和多个待匹配的串，问每个串里出现了多少种模式串。加密方法是把每3bytes加密成按6bits一个对应成4个字符，对应方法题里给了。

分析：除了解密之外，基本是个赤裸裸的AC自动机。这题要注意有多个模式串要进自动机，所以自动机的vis数组要每次清零。

#include <cstdio>

#include <queue>

#include <cstring>

#include <cctype>

using namespace std;



#define D(x) 



const int MAX_CHILD_NUM = 256;

const int MAX_NODE_NUM = 100 * 512 + 10;

const int MAX_LEN = 3000 + 10;



char st[MAX_LEN];

int st2[MAX_LEN];

bool vis[MAX_NODE_NUM];



struct Trie

{

    int next[MAX_NODE_NUM][MAX_CHILD_NUM];

    int fail[MAX_NODE_NUM];

    int count[MAX_NODE_NUM];

    int node_cnt;

    int root;



    void init()

    {

        node_cnt = 0;

        root = newnode();

    }



    int newnode()

    {

        for (int i = 0; i < MAX_CHILD_NUM; i++)

            next[node_cnt][i] = -1;

        count[node_cnt++] = 0;

        return node_cnt - 1;

    }



    int get_id(int a)

    {

        return a;

    }



    void insert(int buf[], int id)

    {

        int now = root;

        for (int i = 0; buf[i] != -1; i++)

        {

            int id = get_id(buf[i]);

            if (next[now][id] == -1)

                next[now][id] = newnode();

            now = next[now][id];

        }

        count[now]++;

    }



    void build()

    {

        queue<int>Q;

        fail[root] = root;

        for (int i = 0; i < MAX_CHILD_NUM; i++)

            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 (int i = 0; i < MAX_CHILD_NUM; i++)

                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(int buf[])

    {

        int now = root;

        int res = 0;

        for (int i = 0; buf[i] != -1; i++)

        {

            now = next[now][get_id(buf[i])];

            int temp = now;

            while (temp != root && !vis[temp])

            {

                res += count[temp];

                 // optimization: prevent from searching this fail chain again.

                //also prevent matching again.

                vis[temp] = true;

                temp = fail[temp];

            }

        }

        return res;

    }



    void debug()

    {

        for(int i = 0;i < node_cnt;i++)

        {

            printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],count[i]);

            for(int j = 0;j < MAX_CHILD_NUM;j++)

                printf("%2d",next[i][j]);

            printf("]\n");

        }

    }

}ac;



int n, m;



int get_value(char ch)

{

    if (isupper(ch))

        return ch - 'A';

    if (islower(ch))

        return ch - 'a' + 26;

    if (isdigit(ch))

        return ch - '0' + 52;

    if (ch == '+')

        return 62;

    return 63;

}



void transform(char *st, int *st2)

{

    int len = strlen(st);

    int len2 = len * 3 / 4;

    for (int i = 0; i < len; i += 4)

    {

        int a = 0;

        for (int j = 0; j < 4; j++)

        {

                a = (a << 6) + get_value(st[i + j]);

            D(printf("**%d\n", a));

        }

        

        for (int j = 2; j >= 0; j--)

        {

            st2[i * 3 / 4 + j] = a % (1 << 8);

            a >>= 8;

            D(printf("**%d\n", st2[i * 3 / 4 + j]));

        }

    }

    while (st[len - 1] == '=')

    {

        len--;

        len2--;

    }

    st2[len2] = -1;

    D(puts("#"));

    for (int i = 0; i < len2; i++)

    {

        D(printf("%d ", st2[i]));

    }

    D(puts(""));

}



void input()

{

    for (int i = 1; i <= n; i++)

    {

        scanf("%s", st);

        transform(st, st2);

        ac.insert(st2, i);

    }

    ac.build();

    scanf("%d", &m);

    for (int i = 0; i < m; i++)

    {

        scanf("%s", st);

        transform(st, st2);

        memset(vis, 0, sizeof(vis));

        printf("%d\n", ac.query(st2));

    }

    puts("");

}



int main()

{



    while (scanf("%d", &n) != EOF)

    {

        ac.init();

        input();

    }

    return 0;

}

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