POJ 2778 DNA Sequence（AC自动机+矩阵快速幂）

题目链接：http://poj.org/problem?id=2778

题意：有m种DNA序列是有疾病的，问有多少种长度为n的DNA序列不包含任何一种有疾病的DNA序列。（仅含A,T,C,G四个字符）

思路：Trie图的状态转移，用矩阵mat[i][j]来表示从结点i到j只走一步有几种走法，那么mat的n次幂就表示从结点i到j走n步有几种走法，题目要求解的就是从头节点走n步且不包含危险结点的走法。

mat = mat^n   ans = (mat[0][0] + mat[0][1] + ... + mat[0][num]) num为结点个数

code：

#include <cstdio>
#include <cstring>
#include <queue>
#include <map>
using namespace std;
const int KIND = 4;
const int MAXN = 110;
const int MOD = 100000;
typedef long long LL;

struct Trie
{
    int next[MAXN][KIND], fail[MAXN];
    bool isExit[MAXN];
    int root, L;
    map<char, int> mp;
    LL mat[MAXN][MAXN];
    LL ret[MAXN][MAXN];
    LL tmp[MAXN][MAXN];
    int create()
    {
        for (int i = 0; i < KIND; ++i)
            next[L][i] = -1;
        isExit[L++] = false;
        return L - 1;
    }
    void init()
    {
        L = 0;
        root = create();
        mp['A'] = 0;
        mp['C'] = 1;
        mp['G'] = 2;
        mp['T'] = 3;
        memset(mat, 0, sizeof(mat));
        memset(ret, 0, sizeof(ret));
    }
    void insert(char str[])
    {
        int now = root;
        int len = strlen(str);
        for (int i = 0; i < len; ++i)
        {
            if (-1 == next[now][mp[str[i]]])
                next[now][mp[str[i]]] = create();
            now = next[now][mp[str[i]]];
        }
        isExit[now] = true;
    }
    void build()
    {
        queue<int>Q;
        for (int i = 0; i < KIND; ++i)
        {
            if (-1 == next[root][i])
                next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        }
        while (!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            if (isExit[fail[now]])
                isExit[now] = true;
            for (int i = 0; i < KIND; ++i)
            {
                if (-1 == next[now][i])
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
            }
        }
    }
    void getMatrix()
    {
        for (int i = 0; i < L; ++i)
        {
            for (int j = 0; j < KIND; ++j)
            {
                if (!isExit[next[i][j]])
                    ++mat[i][next[i][j]];
            }
        }
    }
    void matrixMul(LL mat1[MAXN][MAXN], LL mat2[MAXN][MAXN])
    {
        LL mat3[MAXN][MAXN];
        for (int i = 0; i < L; ++i)
        {
            for (int j = 0; j < L; ++j)
            {
                mat3[i][j] = 0;
                for (int k = 0; k < L; ++k)
                    mat3[i][j] = (mat3[i][j] + mat1[i][k] * mat2[k][j]) % MOD;
            }
        }
        memcpy(mat1, mat3, sizeof(mat3));
    }
    void matrixQuickMod(LL n)
    {
        getMatrix();
        for (int i = 0; i < L; ++i)
        {
            ret[i][i] = 1;
            for (int j = 0; j < L; ++j)
                tmp[i][j] = mat[i][j];
        }
        while (n)
        {
            if (n & 1) matrixMul(ret, tmp);
            matrixMul(tmp, tmp);
            n >>= 1;
        }
    }
};
Trie ac;
char str[15];
int main()
{
    int m;
    LL n;
    while (scanf("%d %lld", &m, &n) != EOF)
    {
        ac.init();
        for (int i = 0; i < m; ++i)
        {
            scanf("%s", str);
            ac.insert(str);
        }
        ac.build();
        ac.matrixQuickMod(n);
        int ans = 0;
        for (int i = 0; i < ac.L; ++i)
            ans = (ans + ac.ret[0][i]) % MOD;
        printf("%d
", ans);
    }
    return 0;
}