POJ - 2778 ~ HDU - 2243 AC自动机+矩阵快速幂

这两题属于AC自动机的第二种套路通过矩阵快速幂求方案数。

题意：给m个病毒字符串，问长度为n的DNA片段有多少种没有包含病毒串的。

根据AC自动机的tire图，我们可以获得一个可达矩阵。

关于这题的tire图详解可以点击这里，往下面翻，这个博主的图对于tire图讲的非常详细。

知道了什么是tire图，理解了tire图后，后面的AC自动机的题目才能写。

AC自动机的灵魂应该就是tire图

然后问题就变成了，得到了一个可达矩阵后，如何求方案数呢？

这个n = 2000000000 这咋办呢？

给定一个有向图，问从A点恰好走k步（允许重复经过边）到达B点的方案数mod p的值

这个是一个关于矩阵快速木的经典问题。

把给定的图转为邻接矩阵，即A(i,j)=1当且仅当存在一条边i->j。

令C=A*A，那么C(i,j)=ΣA(i,k)*A(k,j)，实际上就等于从点i到点j恰好经过2条边的路径数（枚举k为中转点）。

类似地，C*A的第i行第j列就表示从i到j经过3条边的路径数。

同理，如果要求经过k步的路径数，我们只需要二分求出A^k即可。

是不是就是一个裸的矩阵快速幂了。

通过AC自动机得到可达矩阵，然后通过矩阵快速幂求方案数。

 #include <cstdio>
 #include <cstring>
 #include <queue>
 #include <cmath>
 #include <algorithm>
 #include <set>
 #include <iostream>
 #include <map>
 #include <stack>
 #include <string>
 #include <ctime>
 #include <vector>

 #define  pi acos(-1.0)
 #define  eps 1e-9
 #define  fi first
 #define  se second
 #define  rtl   rt<<1
 #define  rtr   rt<<1|1
 #define  bug         printf("******\n")
 #define  mem(a, b)    memset(a,b,sizeof(a))
 #define  name2str(x) #x
 #define  fuck(x)     cout<<#x" = "<<x<<endl
 #define  sf(n)       scanf("%d", &n)
 #define  sff(a, b)    scanf("%d %d", &a, &b)
 #define  sfff(a, b, c) scanf("%d %d %d", &a, &b, &c)
 #define  sffff(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
 #define  pf          printf
 #define  FIN               freopen("../date.txt","r",stdin)
 #define  gcd(a, b)    __gcd(a,b)
 #define  lowbit(x)   x&-x

 using namespace std;
 typedef long long LL;
 typedef unsigned long long ULL;
 const int maxn = 3e3 + ;
 const int maxm = 8e6 + ;
 const int INF = 0x3f3f3f3f;
 const int mod = ;

 struct Matrix {
     int mat[][], n;

     Matrix() {}

     Matrix(int _n) {
         n = _n;
         for (int i = ; i < n; i++)
             for (int j = ; j < n; j++)
                 mat[i][j] = ;
     }

     Matrix operator*(const Matrix &b) const {
         Matrix ret = Matrix(n);
         for (int i = ; i < n; i++)
             for (int j = ; j < n; j++)
                 for (int k = ; k < n; k++) {
                     int tmp = (long long) mat[i][k] * b.mat[k][j] % mod;
                     ret.mat[i][j] = (ret.mat[i][j] + tmp) % mod;
                 }
         return ret;
     }
 };

 Matrix pow_M(Matrix a, int b) {
     Matrix ret = Matrix(a.n);
     for (int i = ; i < ret.n; i++)
         ret.mat[i][i] = ;
     Matrix tmp = a;
     while (b) {
         if (b & )ret = ret * tmp;
         tmp = tmp * tmp;
         b >>= ;
     }
     return ret;
 }

 struct Aho_Corasick {
     int next[][], fail[], End[];
     int root, cnt;

     int newnode() {
         for (int i = ; i < ; i++) next[cnt][i] = -;
         End[cnt++] = ;
         return cnt - ;
     }

     void init() {
         cnt = ;
         root = newnode();
     }

     int get_num(char ch) {
         if (ch == 'A') return ;
         if (ch == 'T') return ;
         if (ch == 'C') return ;
         if (ch == 'G') return ;
     }

     void insert(char buf[]) {
         int len = strlen(buf);
         int now = root;
         for (int i = ; i < len; i++) {
             if (next[now][get_num(buf[i])] == -) next[now][get_num(buf[i])] = newnode();
             now = next[now][get_num(buf[i])];
         }
         End[now]++;
     }

     void build() {
         queue<int> Q;
         fail[root] = root;
         for (int i = ; i < ; i++)
             if (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 (End[fail[now]]) End[now] = ;
             for (int i = ; i < ; 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]);
                 }
         }
     }

     Matrix get_Matrix() {
         Matrix ret = Matrix(cnt);
         for (int i = ; i < cnt; ++i) {
             for (int j = ; j < ; ++j) {
                 if (End[next[i][j]]) continue;
                 ret.mat[i][next[i][j]]++;
             }
         }
         return ret;
     }

     int query(char buf[]) {
         int len = strlen(buf);
         int now = root;
         int res = ;
         for (int i = ; i < len; i++) {
             now = next[now][buf[i] - 'a'];
             int temp = now;
             while (temp != root) {
                 res += End[temp];
                 End[temp] = ;
                 temp = fail[temp];
             }
         }
         return res;
     }

     void debug() {
         for (int i = ; i < cnt; i++) {
             printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
             for (int j = ; j < ; j++) printf("%2d", next[i][j]);
             printf("]\n");
         }
     }
 } ac;

 int n, m;
 char str[maxn];

 int main() {
   //  FIN;
     sff(m, n);
     ac.init();
     for (int i = ; i < m; ++i) {
         scanf("%s", str);
         ac.insert(str);
     }
     ac.build();
     Matrix mat = ac.get_Matrix();
     mat = pow_M(mat, n);
     LL ans = ;
     for (int i = ; i < mat.n; ++i) {
         ans = (ans + mat.mat[][i]) % mod;
     }
     printf("%lld\n", ans);
     return ;
 }

考研路茫茫――单词情结 HDU - 2243

这题和上题题意类似，做法一样。

上题是说长度为n的不包含m个模式串的方案数，这题求的是长度为1~n不包括m个模式串的方案数。

这题就在矩阵上面加上一列全为1的列，这个可以1保存了前面的方案书之和。

如果不理解的话，建议通过手算一下矩阵，去看看这个新加的一列有什么用。

这题是对2^64取模，所以直接用unsigned long long 自动溢出取模即可。

 #include <set>
 #include <map>
 #include <stack>
 #include <queue>
 #include <cmath>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <time.h>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <unordered_map>

 #define  pi acos(-1.0)
 #define  eps 1e-9
 #define  fi first
 #define  se second
 #define  rtl   rt<<1
 #define  rtr   rt<<1|1
 #define  bug               printf("******\n")
 #define  mem(a, b)         memset(a,b,sizeof(a))
 #define  name2str(x)       #x
 #define  fuck(x)           cout<<#x" = "<<x<<endl
 #define  sf(n)             scanf("%d", &n)
 #define  sff(a, b)         scanf("%d %d", &a, &b)
 #define  sfff(a, b, c)     scanf("%d %d %d", &a, &b, &c)
 #define  sffff(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
 #define  pf                printf
 #define  FIN               freopen("../date.txt","r",stdin)
 #define  gcd(a, b)         __gcd(a,b)
 #define  lowbit(x)         x&-x
 #define  IO                iOS::sync_with_stdio(false)

 using namespace std;
 typedef long long LL;
 typedef unsigned long long ULL;
 const int maxn = 1e6 + ;
 const int maxm = 8e6 + ;
 const int INF = 0x3f3f3f3f;
 const int mod = 1e9 + ;

 struct Matrix {
     ULL mat[][], n;

     Matrix() {}

     Matrix(int _n) {
         n = _n;
         for (int i = ; i < n; i++)
             for (int j = ; j < n; j++)
                 mat[i][j] = ;
     }

     Matrix operator*(const Matrix &b) const {
         Matrix ret = Matrix(n);
         for (int i = ; i < n; i++)
             for (int j = ; j < n; j++)
                 for (int k = ; k < n; k++)
                     ret.mat[i][j] = ret.mat[i][j] + mat[i][k] * b.mat[k][j];
         return ret;
     }
 };

 Matrix pow_M(Matrix a, LL b) {
     Matrix ret = Matrix(a.n);
     for (int i = ; i < ret.n; i++)
         ret.mat[i][i] = ;
     Matrix tmp = a;
     while (b) {
         if (b & )ret = ret * tmp;
         tmp = tmp * tmp;
         b >>= ;
     }
     return ret;
 }

 struct Aho_Corasick {
     int next[][], fail[], End[];
     int root, cnt;

     int newnode() {
         for (int i = ; i < ; i++) next[cnt][i] = -;
         End[cnt++] = ;
         return cnt - ;
     }

     void init() {
         cnt = ;
         root = newnode();
     }

     void insert(char buf[]) {
         int len = strlen(buf);
         int now = root;
         for (int i = ; i < len; i++) {
             if (next[now][buf[i] - 'a'] == -) next[now][buf[i] - 'a'] = newnode();
             now = next[now][buf[i] - 'a'];
         }
         End[now]++;
     }

     void build() {
         queue<int> Q;
         fail[root] = root;
         for (int i = ; i < ; i++)
             if (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 (End[fail[now]]) End[now] = ;
             for (int i = ; i < ; 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]);
                 }
         }
     }

     Matrix get_Matrix() {
         Matrix ret = Matrix(cnt+);
         for (int i = ; i < cnt; ++i) {
             for (int j = ; j < ; ++j) {
                 if (!End[next[i][j]]) ret.mat[i][next[i][j]]++;
             }
         }
         for (int i = ; i <= cnt; ++i) ret.mat[i][cnt] = ;
         return ret;
     }

     int query(char buf[]) {
         int len = strlen(buf);
         int now = root;
         int res = ;
         for (int i = ; i < len; i++) {
             now = next[now][buf[i]];
             int temp = now;
             while (temp != root) {
                 res += End[temp];
                 End[temp] = ;
                 temp = fail[temp];
             }
         }
         return res;
     }

     void debug() {
         for (int i = ; i < cnt; i++) {
             printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
             for (int j = ; j < ; j++) printf("%2d", next[i][j]);
             printf("]\n");
         }
     }
 } ac;

 char buf[];
 LL n, m;

 int main() {
    // FIN;
     while (~scanf("%lld%lld", &n, &m)) {
         ac.init();
         for (int i = ; i < n; ++i) {
             scanf("%s", buf);
             ac.insert(buf);
         }
         ac.build();
         Matrix mat = ac.get_Matrix();
         mat = pow_M(mat, m);
         ULL res = , ans = ;
         for (int i = ; i < mat.n; ++i) res += mat.mat[][i];
         Matrix a = Matrix();
         a.mat[][] = , a.mat[][] = a.mat[][] = ;
         a = pow_M(a, m);
         ans = a.mat[][] + a.mat[][];
         ans -= res;
         printf("%llu\n", ans);
     }
     return ;
 }