poj 1625 Censored!(AC自动机+DP+高精度)

题目链接：poj 1625 Censored!

题目大意：给定N，M，K，然后给定一个N字符的字符集和，现在要用这些字符组成一个长度为M的字符串，要求不包

括K个子字符串。

解题思路：AC自动机+DP+高精度。这题恶心的要死，给定的不能匹配字符串里面有负数的字符情况，也算是涨姿势

了，对应每个字符固定偏移128单位。

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

using namespace std;

typedef long long ll;

const int maxn = 505;
const int sigma_size = 256;
const int bw = 128;

struct Aho_Corasick {
    int sz, g[maxn][sigma_size];
    int tag[maxn], fail[maxn], last[maxn];

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

struct bign {
    int len, num[100];

    bign () {
        len = 0;
        memset(num, 0, sizeof(num));
    }
    bign (int number) {*this = number;}
    bign (const char* number) {*this = number;}

    void delzero ();
    void Put ();

    void operator = (int number);
    void operator = (char* number);

    bool operator <  (const bign& b) const;
    bool operator >  (const bign& b) const { return b < *this; }
    bool operator <= (const bign& b) const { return !(b < *this); }
    bool operator >= (const bign& b) const { return !(*this < b); }
    bool operator != (const bign& b) const { return b < *this || *this < b;}
    bool operator == (const bign& b) const { return !(b != *this); }

    void operator ++ ();
    void operator -- ();
    bign operator + (const int& b);
    bign operator + (const bign& b);
    bign operator - (const int& b);
    bign operator - (const bign& b);
    bign operator * (const int& b);
    bign operator * (const bign& b);
    bign operator / (const int& b);
    int operator % (const int& b);
}dp[2][maxn];

int N, M, K, tab[55];
char s[55];

void solve() {
    int n = A.sz;
    for (int i = 0; i < n; i++)
        dp[0][i] = 0;
    dp[0][0] = 1;

    for (int i = 0; i < M; i++) {
        int now = i&1, nxt = !(i&1);
        for (int i = 0; i < n; i++) dp[nxt][i] = 0;

        for (int i = 0; i < n; i++) {
            if (A.tag[i] || A.last[i])
                continue;

            for (int j = 0; j < N; j++) {
                int v = tab[j], u = i;
                while (u && A.g[u][v] == 0)
                    u = A.fail[u];
                u = A.g[u][v];
                dp[nxt][u] = dp[nxt][u] + dp[now][i];
            }
        }
    }

    int now = (M&1);
    bign ans = 0;
    for (int i = 0; i < n; i++) {
        if (A.tag[i] || A.last[i])
            continue;
        ans = ans + dp[now][i];
    }
    ans.Put();
    printf("\n");
}

int main () {
    while (scanf("%d%d%d%*c", &N, &M, &K) == 3) {
        A.init();
        gets(s);
        for (int i = 0; i < N; i++)
            tab[i] = s[i] + bw;

        for (int i = 1; i <= K; i++) {
            gets(s);
            A.insert(s, i);
        }
        A.getFail();
        solve();
    }
    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 + bw;
}

void Aho_Corasick::put(int x, int y) {
}

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

    for (int i = 0; i < n; i++) {
        int v = idx(str[i]);
        if (g[u][v] == 0) {
            tag[sz] = 0;
            memset(g[sz], 0, sizeof(g[sz]));
            g[u][v] = sz++;
        }
        u = g[u][v];
    }
    tag[u] = k;
}

void Aho_Corasick::match(char* str) {
    int n = strlen(str), u = 0;
    for (int i = 0; i < n; i++) {
        int v = idx(str[i]);
        while (u && g[u][v] == 0)
            u = fail[u];

        u = g[u][v];

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

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]];
        }
    }
}

void bign::delzero () {
    while (len && num[len-1] == 0)
        len--;

    if (len == 0) {
        num[len++] = 0;
    }
}

void bign::Put () {
    for (int i = len-1; i >= 0; i--)
        printf("%d", num[i]);
}

void bign::operator = (char* number) {
    len = strlen (number);
    for (int i = 0; i < len; i++)
        num[i] = number[len-i-1] - ‘0‘;

    delzero ();
}

void bign::operator = (int number) {

    len = 0;
    while (number) {
        num[len++] = number%10;
        number /= 10;
    }

    delzero ();
}

bign bign::operator + (const int& b) {
    bign a = b;
    return *this + a;
}

bign bign::operator + (const bign& b) {
    int bignSum = 0;
    bign ans;

    for (int i = 0; i < len || i < b.len; i++) {
        if (i < len) bignSum += num[i];
        if (i < b.len) bignSum += b.num[i];

        ans.num[ans.len++] = bignSum % 10;
        bignSum /= 10;
    }

    while (bignSum) {
        ans.num[ans.len++] = bignSum % 10;
        bignSum /= 10;
    }

    return ans;
}