POJ2778&HDU2243&POJ1625（AC自动机+矩阵/DP）

POJ2778

题意：只有四种字符的字符串（A, C, T and G），有M中字符串不能出现，为长度为n的字符串可以有多少种。

题解：在字符串上有L中状态，所以就有L*A（字符个数）中状态转移。这里自动机的build的hdu2222略有不同。

那一题中通过询问时循环来求she的he，但是如果he不能出现，she就算不是禁止的字符串也不可出现，所以在build的时候就记录所有不能出现的状态。

if (end[ fail[now] ]) end[now]++;

然后用一个矩阵F来记录可以相互到达的状态就OK了。

矩阵可以理解为用长度为1的字符串两个状态可以相互达到，而F=F^n，F[i][j]就是字符串长度为n时状态i到状态j的路径。

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <queue>
#include <set>
using namespace std;

const int N = ;
const int A = ;
const int M = ;
const int MOD = ;

typedef vector<int> vec;
typedef vector<vec> mat;

mat mul(mat &A, mat &B)
{
    mat C(A.size(), vec(B[].size()));
    for (int i = ; i < A.size(); ++i) {
        for (int k = ; k < B.size(); ++k) {
            for (int j = ; j < B[].size(); ++j) {
                C[i][j] = (C[i][j] + (long long)A[i][k] * B[k][j]) % MOD;
            }
        }
    }
    return C;
}

// A^n
mat pow(mat A, int n)
{
    mat B(A.size(), vec(A.size()));
    for (int i = ; i < A.size(); ++i)
        B[i][i] = ;
    while (n > ) {
        if (n & ) B = mul(B, A);
        A = mul(A, A);
        n >>= ;
    }
    return B;
}

struct ACAutomata {

    int next[N][A], fail[N], end[N];
    int root, L;

    int idx(char ch)
    {
        switch(ch) {
            case 'A': return ;
            case 'C': return ;
            case 'T': return ;
            case 'G': return ;
        }
        return -;
    }
    int newNode()
    {
        for (int i = ; i < A; ++i) next[L][i] = -;
        end[L] = ;
        return L++;
    }
    void init()
    {
        L = ;
        root = newNode();
    }
    void insert(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        for (int i = ; i < len; ++i) {
            int ch = idx(buf[i]);
            if (next[now][ch] == -) next[now][ch] = newNode();
            now = next[now][ch];
        }
        end[now]++;
    }
    void build()
    {
        queue<int> Q;
        fail[root] = root;
        for (int i = ; i < A; ++i) {
            if (next[root][i] == -) {
                next[root][i] = root;
            } else {
                fail[ next[root][i] ] = root;
                Q.push( next[root][i] );
            }
        }
        while (Q.size()) {
            int now = Q.front();
            Q.pop();
            if (end[ fail[now] ]) end[now]++; //!!
            for (int i = ; i < A; ++i) {
                if (next[now][i] == -) {
                    next[now][i] = next[ fail[now] ][i];
                } else {
                    fail[ next[now][i] ] = next[ fail[now] ][i];
                    Q.push(next[now][i]);
                }
            }
        }
    }

    int query(int n)
    {
        mat F(L, vec(L));
        for (int i = ; i < L; ++i) {
            for (int j = ; j < L; ++j) {
                F[i][j] = ;
            }
        }
        for (int i = ; i < L; ++i) {
            for (int j = ; j < A; ++j) {
                int nt = next[i][j];
                if (!end[nt]) F[i][nt]++;
            }
        }
        F = pow(F, n);
        int res = ;
        for (int i = ; i < L; ++i) {
            res = (res + F[][i]) % MOD;
        }
        return res;
    }

} ac;

char buf[];
int main()
{
    int m, n;
    while (~scanf("%d%d", &m, &n)) {
        ac.init();
        while (m--) {
            scanf("%s", buf);
            ac.insert(buf);
        }
        ac.build();
        printf("%d\n", ac.query(n));
    }

    return ;
}

HDU 2243上题要求不存在给定字符串，而这题要求存在至少一个。于是方法就是总方法减去不存在的就是答案了。

题目要求是小于等于L的，所以矩阵加一维记录前缀和就好了。

#include <bits/stdc++.h>

using namespace std;

typedef unsigned long long ull;
typedef vector<ull> vec;
typedef vector<vec> mat;
const int N = ;
const int A = ;
const int M = ;

mat mul(mat &A, mat &B)
{
    mat C(A.size(), vec(B[].size()));
    for (int i = ; i < A.size(); ++i) {
        for (int k = ; k < B.size(); ++k) {
            for (int j = ; j < B[].size(); ++j) {
                C[i][j] += A[i][k] * B[k][j];
            }
        }
    }
    return C;
}

mat pow(mat A, ull n)
{
    mat B(A.size(), vec(A.size()));
    for (int i = ; i < A.size(); ++i)
        B[i][i] = ;
    while (n > ) {
        if (n & ) B = mul(B, A);
        A = mul(A, A);
        n >>= ;
    }
    return B;
}

struct ACAutomata {

    int next[N][A], fail[N], end[N];
    int root, L;

    int idx(char ch)
    {
        return ch - 'a';
    }
    int newNode()
    {
        for (int i = ; i < A; ++i) next[L][i] = -;
        end[L] = ;
        return L++;
    }
    void init()
    {
        L = ;
        root = newNode();
    }
    void insert(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        for (int i = ; i < len; ++i) {
            int ch = idx(buf[i]);
            if (next[now][ch] == -) next[now][ch] = newNode();
            now = next[now][ch];
        }
        end[now]++;
    }
    void build()
    {
        queue<int> Q;
        fail[root] = root;
        for (int i = ; i < A; ++i) {
            if (next[root][i] == -) {
                next[root][i] = root;
            } else {
                fail[ next[root][i] ] = root;
                Q.push( next[root][i] );
            }
        }
        while (Q.size()) {
            int now = Q.front();
            Q.pop();
            if (end[ fail[now] ]) end[now]++;
            for (int i = ; i < A; ++i) {
                if (next[now][i] == -) {
                    next[now][i] = next[ fail[now] ][i];
                } else {
                    fail[ next[now][i] ] = next[ fail[now] ][i];
                    Q.push(next[now][i]);
                }
            }
        }
    }

    ull query(ull n)
    {
        mat F(L+, vec(L+));
        for (int i = ; i <= L; ++i) {
            for (int j = ; j <= L; ++j) {
                F[i][j] = ;
            }
        }
        for (int i = ; i <= L; ++i) F[i][L] = ;
        for (int i = ; i < L; ++i) {
            for (int j = ; j < A; ++j) {
                int nt = next[i][j];
                if (!end[nt]) F[i][nt]++;
            }
        }
        F = pow(F, n+);
        return F[][L] - ;
    }

} ac;

char aff[];
int main()
{
    int n;
    ull m;
    while (~scanf("%d%llu", &n, &m)) {
        ac.init();
        while (n--) {
            scanf("%s", aff);
            ac.insert(aff);
        }
        ac.build();
        ull ans1 = ac.query(m);
        mat F(, vec());
        mat S(, vec()); S[][] = ; S[][] = ;
        F[][] = ; F[][] = ;
        F[][] = ;  F[][] = ;
        F = pow(F, m);
        S = mul(F, S);
        ull ans2 = S[][];
        printf("%llu\n", ans2 - ans1);
    }
    return ;
}

POJ1625

同poj2778，不过没有取模操作，需要大数。

大数不可以矩阵快速幂吗？内存不够……

使用dp递推，递推公式也比较简单。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <queue>
#include <map>

using namespace std;

typedef vector<int> vec;
typedef vector<vec> mat;
const int N = ;
const int A = ;

map<char, int> mp;

struct BigInt
{
    const static int mod = ;
    const static int DLEN = ;
    int a[],len;
    BigInt()
    {
        memset(a,,sizeof(a));
        len = ;
    }
    BigInt(int v)
    {
        memset(a,,sizeof(a));
        len = ;
        do
        {
            a[len++] = v%mod;
            v /= mod;
        }while(v);
    }
    BigInt(const char s[])
    {
        memset(a,,sizeof(a));
        int L = strlen(s);
        len = L/DLEN;
        if(L%DLEN)len++;
        int index = ;
        for(int i = L-;i >= ;i -= DLEN)
        {
            int t = ;
            int k = i - DLEN + ;
            if(k < )k = ;
            for(int j = k;j <= i;j++)
                t = t* + s[j] - '';
            a[index++] = t;
        }
    }
    BigInt operator +(const BigInt &b)const
    {
        BigInt res;
        res.len = max(len,b.len);
        for(int i = ;i <= res.len;i++)
            res.a[i] = ;
        for(int i = ;i < res.len;i++)
        {
            res.a[i] += ((i < len)?a[i]:)+((i < b.len)?b.a[i]:);
            res.a[i+] += res.a[i]/mod;
            res.a[i] %= mod;
        }
        if(res.a[res.len] > )res.len++;
        return res;
    }
    BigInt operator *(const BigInt &b)const
    {
        BigInt res;
        for(int i = ; i < len;i++)
        {
            int up = ;
            for(int j = ;j < b.len;j++)
            {
                int temp = a[i]*b.a[j] + res.a[i+j] + up;
                res.a[i+j] = temp%mod;
                up = temp/mod;
            }
            if(up != )
                res.a[i + b.len] = up;
        }
        res.len = len + b.len;
        while(res.a[res.len - ] ==  &&res.len > )res.len--;
        return res;
    }
    void output()
    {
        printf("%d",a[len-]);
        for(int i = len-;i >= ;i--)
            printf("%04d",a[i]);
        printf("\n");
    }
};

struct ACAutomata {

    int next[N][A], fail[N], end[N];
    int root, L;

    int idx(char ch)
    {
        return mp[ch];
    }
    int newNode()
    {
        for (int i = ; i < A; ++i) next[L][i] = -;
        end[L] = ;
        return L++;
    }
    void init()
    {
        L = ;
        root = newNode();
    }
    void insert(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        for (int i = ; i < len; ++i) {
            int ch = idx(buf[i]);
            if (next[now][ch] == -) next[now][ch] = newNode();
            now = next[now][ch];
        }
        end[now]++;
    }
    void build()
    {
        queue<int> Q;
        fail[root] = root;
        for (int i = ; i < A; ++i) {
            if (next[root][i] == -) {
                next[root][i] = root;
            } else {
                fail[ next[root][i] ] = root;
                Q.push( next[root][i] );
            }
        }
        while (Q.size()) {
            int now = Q.front();
            Q.pop();
            if (end[ fail[now] ]) end[now]++; //!!
            for (int i = ; i < A; ++i) {
                if (next[now][i] == -) {
                    next[now][i] = next[ fail[now] ][i];
                } else {
                    fail[ next[now][i] ] = next[ fail[now] ][i];
                    Q.push(next[now][i]);
                }
            }
        }
    }
    mat query(int n)
    {
        mat F(L, vec(L));
        for (int i = ; i < L; ++i) {
            for (unsigned j = ; j < mp.size(); ++j) {
                int nt = next[i][j];
                if (!end[nt]) F[i][nt]++;
            }
        }
        return F;
    }

} ac;

char word[];
BigInt dp[][];
int main()
{
    int n, m, p;
    while (~scanf("%d%d%d", &n, &m, &p)) {
        ac.init();
        getchar();
        gets(word);
        mp.clear();
        for (unsigned i = ; i < strlen(word); ++i) mp[ word[i] ] = i;
        while (p--) {
            gets(word);
            ac.insert(word);
        }
        ac.build();
        mat F = ac.query(m);

        int now = ;
        dp[now][] = ;
        for (int i = ; i < ac.L; ++i) dp[now][i] = ;
        for (int i = ; i <= m; ++i) {
            now ^= ;
            for (int j = ; j < ac.L; ++j) dp[now][j] = ;
            for (int j = ; j < ac.L; ++j)
                for (int k = ; k < ac.L; ++k)
                    if (F[j][k]) dp[now][k] = dp[now][k] + dp[now^][j] * F[j][k];
        }
        BigInt res = ;
        for (int i = ; i < ac.L; ++i) res = res + dp[now][i];
        res.output();
    }
    return ;
}