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。
1 #include <cstdio>
2 #include <cstring>
3 #include <queue>
4 #include <cmath>
5 #include <algorithm>
6 #include <set>
7 #include <iostream>
8 #include <map>
9 #include <stack>
10 #include <string>
11 #include <ctime>
12 #include <vector>
13
14 #define pi acos(-1.0)
15 #define eps 1e-9
16 #define fi first
17 #define se second
18 #define rtl rt<<1
19 #define rtr rt<<1|1
20 #define bug printf("******\n")
21 #define mem(a, b) memset(a,b,sizeof(a))
22 #define name2str(x) #x
23 #define fuck(x) cout<<#x" = "<<x<<endl
24 #define sf(n) scanf("%d", &n)
25 #define sff(a, b) scanf("%d %d", &a, &b)
26 #define sfff(a, b, c) scanf("%d %d %d", &a, &b, &c)
27 #define sffff(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
28 #define pf printf
29 #define FIN freopen("../date.txt","r",stdin)
30 #define gcd(a, b) __gcd(a,b)
31 #define lowbit(x) x&-x
32
33 using namespace std;
34 typedef long long LL;
35 typedef unsigned long long ULL;
36 const int maxn = 3e3 + 7;
37 const int maxm = 8e6 + 10;
38 const int INF = 0x3f3f3f3f;
39 const int mod = 100000;
40
41 struct Matrix {
42 int mat[110][110], n;
43
44 Matrix() {}
45
46 Matrix(int _n) {
47 n = _n;
48 for (int i = 0; i < n; i++)
49 for (int j = 0; j < n; j++)
50 mat[i][j] = 0;
51 }
52
53 Matrix operator*(const Matrix &b) const {
54 Matrix ret = Matrix(n);
55 for (int i = 0; i < n; i++)
56 for (int j = 0; j < n; j++)
57 for (int k = 0; k < n; k++) {
58 int tmp = (long long) mat[i][k] * b.mat[k][j] % mod;
59 ret.mat[i][j] = (ret.mat[i][j] + tmp) % mod;
60 }
61 return ret;
62 }
63 };
64
65 Matrix pow_M(Matrix a, int b) {
66 Matrix ret = Matrix(a.n);
67 for (int i = 0; i < ret.n; i++)
68 ret.mat[i][i] = 1;
69 Matrix tmp = a;
70 while (b) {
71 if (b & 1)ret = ret * tmp;
72 tmp = tmp * tmp;
73 b >>= 1;
74 }
75 return ret;
76 }
77
78 struct Aho_Corasick {
79 int next[50010][4], fail[50010], End[50010];
80 int root, cnt;
81
82 int newnode() {
83 for (int i = 0; i < 4; i++) next[cnt][i] = -1;
84 End[cnt++] = 0;
85 return cnt - 1;
86 }
87
88 void init() {
89 cnt = 0;
90 root = newnode();
91 }
92
93 int get_num(char ch) {
94 if (ch == ‘A‘) return 0;
95 if (ch == ‘T‘) return 1;
96 if (ch == ‘C‘) return 2;
97 if (ch == ‘G‘) return 3;
98 }
99
100 void insert(char buf[]) {
101 int len = strlen(buf);
102 int now = root;
103 for (int i = 0; i < len; i++) {
104 if (next[now][get_num(buf[i])] == -1) next[now][get_num(buf[i])] = newnode();
105 now = next[now][get_num(buf[i])];
106 }
107 End[now]++;
108 }
109
110 void build() {
111 queue<int> Q;
112 fail[root] = root;
113 for (int i = 0; i < 4; i++)
114 if (next[root][i] == -1) next[root][i] = root;
115 else {
116 fail[next[root][i]] = root;
117 Q.push(next[root][i]);
118 }
119 while (!Q.empty()) {
120 int now = Q.front();
121 Q.pop();
122 if (End[fail[now]]) End[now] = 1;
123 for (int i = 0; i < 4; i++)
124 if (next[now][i] == -1) next[now][i] = next[fail[now]][i];
125 else {
126 fail[next[now][i]] = next[fail[now]][i];
127 Q.push(next[now][i]);
128 }
129 }
130 }
131
132 Matrix get_Matrix() {
133 Matrix ret = Matrix(cnt);
134 for (int i = 0; i < cnt; ++i) {
135 for (int j = 0; j < 4; ++j) {
136 if (End[next[i][j]]) continue;
137 ret.mat[i][next[i][j]]++;
138 }
139 }
140 return ret;
141 }
142
143 int query(char buf[]) {
144 int len = strlen(buf);
145 int now = root;
146 int res = 0;
147 for (int i = 0; i < len; i++) {
148 now = next[now][buf[i] - ‘a‘];
149 int temp = now;
150 while (temp != root) {
151 res += End[temp];
152 End[temp] = 0;
153 temp = fail[temp];
154 }
155 }
156 return res;
157 }
158
159 void debug() {
160 for (int i = 0; i < cnt; i++) {
161 printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
162 for (int j = 0; j < 26; j++) printf("%2d", next[i][j]);
163 printf("]\n");
164 }
165 }
166 } ac;
167
168 int n, m;
169 char str[maxn];
170
171 int main() {
172 // FIN;
173 sff(m, n);
174 ac.init();
175 for (int i = 0; i < m; ++i) {
176 scanf("%s", str);
177 ac.insert(str);
178 }
179 ac.build();
180 Matrix mat = ac.get_Matrix();
181 mat = pow_M(mat, n);
182 LL ans = 0;
183 for (int i = 0; i < mat.n; ++i) {
184 ans = (ans + mat.mat[0][i]) % mod;
185 }
186 printf("%lld\n", ans);
187 return 0;
188 }
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。
1 #include <set>
2 #include <map>
3 #include <stack>
4 #include <queue>
5 #include <cmath>
6 #include <cstdio>
7 #include <string>
8 #include <vector>
9 #include <time.h>
10 #include <cstring>
11 #include <iostream>
12 #include <algorithm>
13 #include <unordered_map>
14
15 #define pi acos(-1.0)
16 #define eps 1e-9
17 #define fi first
18 #define se second
19 #define rtl rt<<1
20 #define rtr rt<<1|1
21 #define bug printf("******\n")
22 #define mem(a, b) memset(a,b,sizeof(a))
23 #define name2str(x) #x
24 #define fuck(x) cout<<#x" = "<<x<<endl
25 #define sf(n) scanf("%d", &n)
26 #define sff(a, b) scanf("%d %d", &a, &b)
27 #define sfff(a, b, c) scanf("%d %d %d", &a, &b, &c)
28 #define sffff(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
29 #define pf printf
30 #define FIN freopen("../date.txt","r",stdin)
31 #define gcd(a, b) __gcd(a,b)
32 #define lowbit(x) x&-x
33 #define IO iOS::sync_with_stdio(false)
34
35
36 using namespace std;
37 typedef long long LL;
38 typedef unsigned long long ULL;
39 const int maxn = 1e6 + 7;
40 const int maxm = 8e6 + 10;
41 const int INF = 0x3f3f3f3f;
42 const int mod = 1e9 + 7;
43
44 struct Matrix {
45 ULL mat[110][110], n;
46
47 Matrix() {}
48
49 Matrix(int _n) {
50 n = _n;
51 for (int i = 0; i < n; i++)
52 for (int j = 0; j < n; j++)
53 mat[i][j] = 0;
54 }
55
56 Matrix operator*(const Matrix &b) const {
57 Matrix ret = Matrix(n);
58 for (int i = 0; i < n; i++)
59 for (int j = 0; j < n; j++)
60 for (int k = 0; k < n; k++)
61 ret.mat[i][j] = ret.mat[i][j] + mat[i][k] * b.mat[k][j];
62 return ret;
63 }
64 };
65
66 Matrix pow_M(Matrix a, LL b) {
67 Matrix ret = Matrix(a.n);
68 for (int i = 0; i < ret.n; i++)
69 ret.mat[i][i] = 1;
70 Matrix tmp = a;
71 while (b) {
72 if (b & 1)ret = ret * tmp;
73 tmp = tmp * tmp;
74 b >>= 1;
75 }
76 return ret;
77 }
78
79 struct Aho_Corasick {
80 int next[10010][26], fail[10010], End[10010];
81 int root, cnt;
82
83 int newnode() {
84 for (int i = 0; i < 26; i++) next[cnt][i] = -1;
85 End[cnt++] = 0;
86 return cnt - 1;
87 }
88
89 void init() {
90 cnt = 0;
91 root = newnode();
92 }
93
94 void insert(char buf[]) {
95 int len = strlen(buf);
96 int now = root;
97 for (int i = 0; i < len; i++) {
98 if (next[now][buf[i] - ‘a‘] == -1) next[now][buf[i] - ‘a‘] = newnode();
99 now = next[now][buf[i] - ‘a‘];
100 }
101 End[now]++;
102 }
103
104 void build() {
105 queue<int> Q;
106 fail[root] = root;
107 for (int i = 0; i < 26; i++)
108 if (next[root][i] == -1) next[root][i] = root;
109 else {
110 fail[next[root][i]] = root;
111 Q.push(next[root][i]);
112 }
113 while (!Q.empty()) {
114 int now = Q.front();
115 Q.pop();
116 if (End[fail[now]]) End[now] = 1;
117 for (int i = 0; i < 26; i++)
118 if (next[now][i] == -1) next[now][i] = next[fail[now]][i];
119 else {
120 fail[next[now][i]] = next[fail[now]][i];
121 Q.push(next[now][i]);
122 }
123 }
124 }
125
126 Matrix get_Matrix() {
127 Matrix ret = Matrix(cnt+1);
128 for (int i = 0; i < cnt; ++i) {
129 for (int j = 0; j < 26; ++j) {
130 if (!End[next[i][j]]) ret.mat[i][next[i][j]]++;
131 }
132 }
133 for (int i = 0; i <= cnt; ++i) ret.mat[i][cnt] = 1;
134 return ret;
135 }
136
137 int query(char buf[]) {
138 int len = strlen(buf);
139 int now = root;
140 int res = 0;
141 for (int i = 0; i < len; i++) {
142 now = next[now][buf[i]];
143 int temp = now;
144 while (temp != root) {
145 res += End[temp];
146 End[temp] = 0;
147 temp = fail[temp];
148 }
149 }
150 return res;
151 }
152
153 void debug() {
154 for (int i = 0; i < cnt; i++) {
155 printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
156 for (int j = 0; j < 26; j++) printf("%2d", next[i][j]);
157 printf("]\n");
158 }
159 }
160 } ac;
161
162
163
164 char buf[1000010];
165 LL n, m;
166
167 int main() {
168 // FIN;
169 while (~scanf("%lld%lld", &n, &m)) {
170 ac.init();
171 for (int i = 0; i < n; ++i) {
172 scanf("%s", buf);
173 ac.insert(buf);
174 }
175 ac.build();
176 Matrix mat = ac.get_Matrix();
177 mat = pow_M(mat, m);
178 ULL res = 0, ans = 0;
179 for (int i = 0; i < mat.n; ++i) res += mat.mat[0][i];
180 Matrix a = Matrix(2);
181 a.mat[0][0] = 26, a.mat[1][0] = a.mat[1][1] = 1;
182 a = pow_M(a, m);
183 ans = a.mat[0][0] + a.mat[1][0];
184 ans -= res;
185 printf("%llu\n", ans);
186 }
187 return 0;
188 }
原文地址：https://www.cnblogs.com/qldabiaoge/p/11375935.html
时间： 2024-07-28 14:23:04