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 = 150;
const int A = 4;
const int M = 15;
const int MOD = 100000;

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

mat mul(mat &A, mat &B)
{
    mat C(A.size(), vec(B[0].size()));
    for (int i = 0; i < A.size(); ++i) {
        for (int k = 0; k < B.size(); ++k) {
            for (int j = 0; j < B[0].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 = 0; i < A.size(); ++i)
        B[i][i] = 1;
    while (n > 0) {
        if (n & 1) B = mul(B, A);
        A = mul(A, A);
        n >>= 1;
    }
    return B;
}

struct ACAutomata {

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

    int idx(char ch)
    {
        switch(ch) {
            case 'A': return 0;
            case 'C': return 1;
            case 'T': return 2;
            case 'G': return 3;
        }
        return -1;
    }
    int newNode()
    {
        for (int i = 0; i < A; ++i) next[L][i] = -1;
        end[L] = 0;
        return L++;
    }
    void init()
    {
        L = 0;
        root = newNode();
    }
    void insert(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        for (int i = 0; i < len; ++i) {
            int ch = idx(buf[i]);
            if (next[now][ch] == -1) next[now][ch] = newNode();
            now = next[now][ch];
        }
        end[now]++;
    }
    void build()
    {
        queue<int> Q;
        fail[root] = root;
        for (int i = 0; i < A; ++i) {
            if (next[root][i] == -1) {
                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 = 0; i < A; ++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 n)
    {
        mat F(L, vec(L));
        for (int i = 0; i < L; ++i) {
            for (int j = 0; j < L; ++j) {
                F[i][j] = 0;
            }
        }
        for (int i = 0; i < L; ++i) {
            for (int j = 0; j < A; ++j) {
                int nt = next[i][j];
                if (!end[nt]) F[i][nt]++;
            }
        }
        F = pow(F, n);
        int res = 0;
        for (int i = 0; i < L; ++i) {
            res = (res + F[0][i]) % MOD;
        }
        return res;
    }

} ac;

char buf[20];
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
", ac.query(n));
    }

    return 0;
}

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 = 150;
const int A = 26;
const int M = 15;

mat mul(mat &A, mat &B)
{
    mat C(A.size(), vec(B[0].size()));
    for (int i = 0; i < A.size(); ++i) {
        for (int k = 0; k < B.size(); ++k) {
            for (int j = 0; j < B[0].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 = 0; i < A.size(); ++i)
        B[i][i] = 1;
    while (n > 0) {
        if (n & 1) B = mul(B, A);
        A = mul(A, A);
        n >>= 1;
    }
    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 = 0; i < A; ++i) next[L][i] = -1;
        end[L] = 0;
        return L++;
    }
    void init()
    {
        L = 0;
        root = newNode();
    }
    void insert(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        for (int i = 0; i < len; ++i) {
            int ch = idx(buf[i]);
            if (next[now][ch] == -1) next[now][ch] = newNode();
            now = next[now][ch];
        }
        end[now]++;
    }
    void build()
    {
        queue<int> Q;
        fail[root] = root;
        for (int i = 0; i < A; ++i) {
            if (next[root][i] == -1) {
                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 = 0; i < A; ++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]);
                }
            }
        }
    }

    ull query(ull n)
    {
        mat F(L+1, vec(L+1));
        for (int i = 0; i <= L; ++i) {
            for (int j = 0; j <= L; ++j) {
                F[i][j] = 0;
            }
        }
        for (int i = 0; i <= L; ++i) F[i][L] = 1;
        for (int i = 0; i < L; ++i) {
            for (int j = 0; j < A; ++j) {
                int nt = next[i][j];
                if (!end[nt]) F[i][nt]++;
            }
        }
        F = pow(F, n+1);
        return F[0][L] - 1;
    }

} ac;

char aff[10];
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(2, vec(2));
        mat S(2, vec(1)); S[0][0] = 26; S[1][0] = 0;
        F[0][0] = 26; F[0][1] = 0;
        F[1][0] = 1;  F[1][1] = 1;
        F = pow(F, m);
        S = mul(F, S);
        ull ans2 = S[1][0];
        printf("%llu
", ans2 - ans1);
    }
    return 0;
}

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 = 110;
const int A = 256;

map<char, int> mp;

struct BigInt
{
    const static int mod = 10000;
    const static int DLEN = 4;
    int a[600],len;
    BigInt()
    {
        memset(a,0,sizeof(a));
        len = 1;
    }
    BigInt(int v)
    {
        memset(a,0,sizeof(a));
        len = 0;
        do
        {
            a[len++] = v%mod;
            v /= mod;
        }while(v);
    }
    BigInt(const char s[])
    {
        memset(a,0,sizeof(a));
        int L = strlen(s);
        len = L/DLEN;
        if(L%DLEN)len++;
        int index = 0;
        for(int i = L-1;i >= 0;i -= DLEN)
        {
            int t = 0;
            int k = i - DLEN + 1;
            if(k < 0)k = 0;
            for(int j = k;j <= i;j++)
                t = t*10 + s[j] - '0';
            a[index++] = t;
        }
    }
    BigInt operator +(const BigInt &b)const
    {
        BigInt res;
        res.len = max(len,b.len);
        for(int i = 0;i <= res.len;i++)
            res.a[i] = 0;
        for(int i = 0;i < res.len;i++)
        {
            res.a[i] += ((i < len)?a[i]:0)+((i < b.len)?b.a[i]:0);
            res.a[i+1] += res.a[i]/mod;
            res.a[i] %= mod;
        }
        if(res.a[res.len] > 0)res.len++;
        return res;
    }
    BigInt operator *(const BigInt &b)const
    {
        BigInt res;
        for(int i = 0; i < len;i++)
        {
            int up = 0;
            for(int j = 0;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 != 0)
                res.a[i + b.len] = up;
        }
        res.len = len + b.len;
        while(res.a[res.len - 1] == 0 &&res.len > 1)res.len--;
        return res;
    }
    void output()
    {
        printf("%d",a[len-1]);
        for(int i = len-2;i >=0 ;i--)
            printf("%04d",a[i]);
        printf("
");
    }
};


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 = 0; i < A; ++i) next[L][i] = -1;
        end[L] = 0;
        return L++;
    }
    void init()
    {
        L = 0;
        root = newNode();
    }
    void insert(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        for (int i = 0; i < len; ++i) {
            int ch = idx(buf[i]);
            if (next[now][ch] == -1) next[now][ch] = newNode();
            now = next[now][ch];
        }
        end[now]++;
    }
    void build()
    {
        queue<int> Q;
        fail[root] = root;
        for (int i = 0; i < A; ++i) {
            if (next[root][i] == -1) {
                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 = 0; i < A; ++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]);
                }
            }
        }
    }
    mat query(int n)
    {
        mat F(L, vec(L));
        for (int i = 0; i < L; ++i) {
            for (unsigned j = 0; j < mp.size(); ++j) {
                int nt = next[i][j];
                if (!end[nt]) F[i][nt]++;
            }
        }
        return F;
    }

} ac;

char word[100];
BigInt dp[2][110];
int main()
{
    int n, m, p;
    while (~scanf("%d%d%d", &n, &m, &p)) {
        ac.init();
        getchar();
        gets(word);
        mp.clear();
        for (unsigned i = 0; i < strlen(word); ++i) mp[ word[i] ] = i;
        while (p--) {
            gets(word);
            ac.insert(word);
        }
        ac.build();
        mat F = ac.query(m);

        int now = 0;
        dp[now][0] = 1;
        for (int i = 1; i < ac.L; ++i) dp[now][i] = 0;
        for (int i = 1; i <= m; ++i) {
            now ^= 1;
            for (int j = 0; j < ac.L; ++j) dp[now][j] = 0;
            for (int j = 0; j < ac.L; ++j)
                for (int k = 0; k < ac.L; ++k)
                    if (F[j][k]) dp[now][k] = dp[now][k] + dp[now^1][j] * F[j][k];
        }
        BigInt res = 0;
        for (int i = 0; i < ac.L; ++i) res = res + dp[now][i];
        res.output();
    }
    return 0;
}