HDU 2243 考研路茫茫――单词情结 AC自动机 + 矩阵快速幂

题目大意:

就是现在给出一些单词的词根,问长度不超过L ( 1 <= L <  2^31) 的只含小写英文字母的串中至少包含一个词根的串有多少个,输出结果对2^64取模输出


大致思路:

这道题和POJ 2778很像(题解戳这里),但是这题需要求得是含有词根的串的数量,很容易想到先求出不包含的个数,然后用全部的减去它即可,这个地方由于是长度不超过L,那么相对于POJ 2778来说需要求矩阵 A的 1~L次方的第一行的和,这里我用的是二分递归来求得等比矩阵和,但是容易RuntimeError(Stack_Overflow)后来减小了矩阵类的数组就过了

这里应该有更优的解法, 不过我还是就用了递归求矩阵的和过了

另外算26 + 26^2 + ... +26^L的时候也是构造的矩阵算的..

对于2^64取模可以利用unsigned long long 的溢出来解决,也就没什么好说的


代码如下:

Result  :  Accepted     Memory  :  1696 KB     Time  :  421 ms

/*
 * Author: Gatevin
 * Created Time:  2014/11/19 17:42:08
 * File Name: Kagome.cpp
 */
#include
#include
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using namespace std;
const double eps(1e-8);
typedef long long lint;

int n, L, bound;

struct Matrix
{
    unsigned long long a[26][26];
    Matrix()
    {
        for(int i = 0; i < 26; i++)
            for(int j = 0; j < 26; j++)
                a[i][j] = i == j ? 1 : 0;
        return;
    }
    unsigned long long* operator[](int i)
    {
        return a[i];
    }
};

Matrix operator * (Matrix m1, Matrix m2)
{
    Matrix rm;
    for(int i = 0; i < bound; i++)
        for(int j = 0; j < bound; j++)
        {
            rm[i][j] = 0;
            for(int k = 0; k < bound; k++)
            {
                rm[i][j] += m1[i][k]*m2[k][j];
            }
        }
    return rm;
}

Matrix operator + (Matrix m1, Matrix m2)
{
    Matrix rm;
    for(int i = 0; i < bound; i++)
        for(int j = 0; j < bound; j++)
            rm[i][j] = m1[i][j] + m2[i][j];
    return rm;
}

Matrix quick(Matrix base, int pow)//快速幂
{
    Matrix I;
    while(pow)
    {
        if(pow & 1)
            I = I * base;
        base = base * base;
        pow >>= 1;
    }
    return I;
}


/*
 * 这个地方像我这么做容易爆栈,刚开始交的时候爆栈了好几次
 * 应该有更好的方法来代替
 */
Matrix Geo(Matrix base, int pow)//矩阵二分求和base + base^2 + base^3 + ... + base^pow
{
    if(pow == 1) return base;
    Matrix I;
    if(pow & 1) return Geo(base, (pow - 1) >> 1) * ( I + quick(base, (pow - 1) >> 1)) + quick(base, pow);
    return Geo(base, pow >> 1)*( I + quick(base, pow >> 1));
}

struct Trie
{
    int next[6*6][26], fail[6*6];
    bool end[6*6], vis[6*6];
    int L, root;
    int newnode()
    {
        for(int i = 0; i < 26; i++)
            next[L][i] = -1;
        end[L++] = 0;
        return L - 1;
    }
    void init()
    {
        L = 0;
        root = newnode();
        return;
    }
    void insert(char *s)
    {
        int now = root;
        for(; *s; s++)
        {
            if(next[now][*s - 'a'] == -1)
                next[now][*s - 'a'] = newnode();
            now = next[now][*s - 'a'];
        }
        end[now] = 1;
        return;
    }
    void build()//建立状态转移图
    {
        fail[root] = root;
        queue  Q;
        Q.push(root);
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            if(end[fail[now]]) end[now] = 1;
            for(int i = 0; i < 26; i++)
                if(next[now][i] == -1)
                    next[now][i] = now == root ? root : next[fail[now]][i];
                else
                {
                    fail[next[now][i]] = now == root ? root : next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
        return;
    }
    Matrix matrix()//BFS计算矩阵
    {
        unsigned long long m[26][26];
        memset(m, 0, sizeof(m));
        memset(vis, 0, sizeof(vis));
        queue  Q;
        Q.push(root);
        vis[root] = 1;
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            for(int i = 0; i < 26; i++)
                if(!end[next[now][i]])
                {
                    m[now][next[now][i]]++;
                    if(!vis[next[now][i]])
                    {
                        Q.push(next[now][i]);
                        vis[next[now][i]] = 1;
                    }
                }
        }
        Matrix R;
        vector  v;
        for(int i = 0; i < L; i++)
            if(vis[i]) v.push_back(i);
        for(unsigned int i = 0; i < v.size(); i++)
            for(unsigned int j = 0; j < v.size(); j++)
                R[i][j] = m[v[i]][v[j]];
        bound = v.size();
        return R;
    }
};

Trie AC;

int main()
{
    char ts[8];
    while(~scanf("%d %d", &n, &L))
    {
        AC.init();
        while(n--)
        {
            scanf("%s", ts);
            AC.insert(ts);
        }
        AC.build();
        Matrix A = AC.matrix();
        Matrix F = Geo(A, L);
        unsigned long long ans1 = 0;
        for(int i = 0; i < bound; i++)
            ans1 += F[0][i];
        Matrix N;
        N[0][0] = 1; N[0][1] = 1;
        N[1][0] = 0; N[1][1] = 26;
        bound = 2;
        Matrix ALL = quick(N, L);//矩阵快速幂求等比数列26 + 26^2 + ... + 26^n的值
        unsigned long long ans2 = ALL[0][1] + ALL[1][1] - 1;
        unsigned long long ans = ans2 - ans1;
        printf("%I64u\n", ans);
    }
    return 0;
}


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