题意：给一些字符串的集合S和整数n，求满足
- 长度为n
- 只含charset = {'A'、'T‘、'G'、'C'}包含的字符
- 不包含S中任一字符串
的字符串的种类数。
思路：首先对S建立ac自动机，考虑向ac自动机中的每种状态后加charset中的字符，如果终态不为“接受状态”，也就是不与S中的任一字符串匹配，则将这次转移记为有效，方法数加1。这样可以建立状态之间的转移矩阵D，表示由一个状态接受1个字符后的方案数，D自乘n次，就得到了任一状态接受n个字符形成的不同字符串种类数，其中从“0”到“i”的方案都要加到答案里面。
另外，这题的模比较有意思，为100000，既不大也不小，或者说是个比较猥琐的数。对于100*100的矩阵，这样取模当然是吃不消的，方法是对每个点算完后一次性取模。时间由800ms→125ms。
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#pragma comment(linker, "/STACK:10240000") #include <map> #include <set> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include <vector> #include <cstdio> #include <string> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define X first #define Y second #define pb push_back #define mp make_pair #define all(a) (a).begin(), (a).end() #define fillchar(a, x) memset(a, x, sizeof(a)) #define fillarray(a, b) memcpy(a, b, sizeof(a)) typedef long long ll; typedef pair<int, int> pii; typedef unsigned long long ull; #ifndef ONLINE_JUDGE namespace Debug { void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);} void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1; while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> void print(const T t){cout<<t<<endl;}template<typename F,typename...R> void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} } #endif // ONLINE_JUDGE template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} const double PI = acos(-1.0); const int INF = 0x3f3f3f3f; const double EPS = 1e-14; /* -------------------------------------------------------------------------------- */ const int maxn = 107; const int mod = 100000; int N; struct Matrix { ll a[maxn][maxn]; Matrix() { for (int i = 0; i < N; i ++) { for (int j = 0; j < N; j ++) { a[i][j] = 0; } } } static Matrix unit() { Matrix ans; for (int i = 0; i < N; i ++) ans.a[i][i] = 1; return ans; } Matrix &operator * (const Matrix &that) const { static Matrix ans; for (int i = 0; i < N; i ++) { for (int j = 0; j < N; j ++) { ans.a[i][j] = 0; for (int k = 0; k < N; k ++) { ans.a[i][j] += a[i][k] * that.a[k][j]; } ans.a[i][j] %= mod; } } return ans; } static Matrix power(Matrix a, int n) { Matrix ans = unit(), buf = a; while (n) { if (n & 1) ans = ans * buf; buf = buf * buf; n >>= 1; } return ans; } }; struct AC_automaton { const static int SZ = 4; int f[maxn], last[maxn]; bool val[maxn]; int ch[maxn][SZ]; int sz; int idx(char ch) { if (ch == 'A') return 0; if (ch == 'T') return 1; if (ch == 'G') return 2; if (ch == 'C') return 3; } void init() { sz = 1; memset(ch[0], 0, sizeof(ch[0])); } void insert(char s[]) { int u = 0; for (int i = 0; s[i]; i ++) { int c = idx(s[i]); if (!ch[u][c]) { memset(ch[sz], 0, sizeof(ch[sz])); val[sz] = 0; ch[u][c] = sz ++; } u = ch[u][c]; } val[u] = true; } void build() { queue<int> Q; f[0] = 0; for (int c = 0; c < SZ; c ++) { int u = ch[0][c]; if (u) { f[u] = 0; Q.push(u); last[u] = 0; } } while (!Q.empty()) { int r = Q.front(); Q.pop(); for (int c = 0; c < SZ; c ++) { int u = ch[r][c]; if (!u) { ch[r][c] = ch[f[r]][c]; continue; } Q.push(u); int v = f[r]; while (v && !ch[v][c]) v = f[v]; f[u] = ch[v][c]; last[u] = val[f[u]]? f[u] : last[f[u]]; } } } void solve(int m) { N = sz; Matrix ans; //Debug::print(sz); for (int i = 0; i < sz; i ++) { for (int j = 0; j < 4; j ++) { if (!val[ch[i][j]] && !last[ch[i][j]]) { ans.a[i][ch[i][j]] ++; } } } ans = Matrix::power(ans, m); int rst = 0; for (int i = 0; i < sz; i ++) { rst = (rst + ans.a[0][i]) % mod; } printf("%d ", rst); } }; AC_automaton ac; int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGE int n, m; char s[20]; while (cin >> n >> m) { ac.init(); for (int i = 0; i < n; i ++) { scanf("%s", s); ac.insert(s); } ac.build(); ac.solve(m); } return 0; } |