标签：hdu continue empty any math 理解 ctime map def
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 }
POJ - 2778 ~ HDU - 2243 AC自动机+矩阵快速幂
标签：hdu continue empty any math 理解 ctime map def
原文地址：https://www.cnblogs.com/qldabiaoge/p/11375935.html
