DNA Sequence
Time Limit: 1000ms
Memory Limit: 65536KB
This problem will be judged on PKU. Original ID: 2778
64-bit integer IO format: %lld Java class name: Main
It‘s well known that DNA Sequence is a sequence only contains A, C, T and G, and it‘s very useful to analyze a segment of DNA Sequence，For example, if a animal‘s DNA sequence contains segment ATC then it may mean that the animal may have a genetic disease. Until now scientists have found several those segments, the problem is how many kinds of DNA sequences of a species don‘t contain those segments.
Suppose that DNA sequences of a species is a sequence that consist of A, C, T and G，and the length of sequences is a given integer n.
Input
First line contains two integer m (0 <= m <= 10), n (1 <= n <=2000000000). Here, m is the number of genetic disease segment, and n is the length of sequences.
Next m lines each line contain a DNA genetic disease segment, and length of these segments is not larger than 10.
Output
An integer, the number of DNA sequences, mod 100000.
Sample Input
4 3 AT AC AG AA
Sample Output
36
Source
解题：真TMD的卡时间，奶奶个熊
1 #include <iostream>
2 #include <stdio.h>
3 #include <cstring>
4 #include <queue>
5 using namespace std;
6 typedef long long LL;
7 const int mod = 100000;
8 const int maxn = 80;
9 int id[256];
10 struct Matrix{
11 int n,m[maxn][maxn];
12 void init(int sz){
13 n = sz;
14 memset(m,0,sizeof m);
15 }
16 Matrix(int sz){
17 init(sz);
18 }
19 void setOne(){
20 for(int i = 0; i < n; ++i) m[i][i] = 1;
21 }
22 Matrix operator*(const Matrix &rhs){
23 Matrix ret(n);
24 for(int k = 0; k < n; ++k)
25 for(int i = 0; i < n; ++i)
26 for(int j = 0; j < n; ++j)
27 ret.m[i][j] = (ret.m[i][j] + (LL)m[i][k]*rhs.m[k][j]%mod)%mod;
28 return ret;
29 }
30 void out(){
31 for(int i = 0; i < n; ++i){
32 for(int j = 0; j < n; ++j)
33 cout<<m[i][j]<<" ";
34 cout<<endl;
35 }
36 }
37 };
38 void quickPow(Matrix &base,Matrix &ret,int index){
39 ret.setOne();
40 while(index){
41 if(index&1) ret = ret*base;
42 index >>= 1;
43 base = base*base;
44 }
45 }
46 struct Trie{
47 int ch[maxn][4],fail[maxn],cnt[maxn],tot;
48 void init(){
49 tot = 0;
50 newnode();
51 }
52 int newnode(){
53 memset(ch[tot],0,sizeof ch[tot]);
54 fail[tot] = cnt[tot] = 0;
55 return tot++;
56 }
57 void insert(char *str,int root = 0){
58 for(int i = 0; str[i]; ++i){
59 int &x = ch[root][id[str[i]]];
60 if(!x) x = newnode();
61 root = x;
62 }
63 ++cnt[root];
64 }
65 void build(int root = 0){
66 queue<int>q;
67 for(int i = 0; i < 4; ++i)
68 if(ch[root][i]) q.push(ch[root][i]);
69 while(!q.empty()){
70 root = q.front();
71 q.pop();
72 cnt[root] += cnt[fail[root]];
73 for(int i = 0; i < 4; ++i){
74 int &x = ch[root][i];
75 if(x){
76 fail[x] = ch[fail[root]][i];
77 q.push(x);
78 }else x = ch[fail[root]][i];
79 }
80 }
81 }
82 int solve(int m){
83 Matrix a(tot),b(tot);
84 for(int i = 0; i < tot; ++i){
85 if(cnt[i]) continue;
86 for(int j = 0; j < 4; ++j){
87 int x = ch[i][j];
88 if(cnt[x]) continue;
89 ++a.m[i][x];
90 }
91 }
92 quickPow(a,b,m);
93 int ret = 0;
94 for(int i = 0; i < tot; ++i)
95 ret = (ret + b.m[0][i])%mod;
96 return ret;
97 }
98 }ac;
99 int main(){
100 for(int i = 0; i < 4; ++i) id["ATCG"[i]] = i;
101 char str[maxn];
102 int n,m;
103 while(~scanf("%d%d",&n,&m)){
104 ac.init();
105 for(int i = 0; i < n; ++i){
106 scanf("%s",str);
107 ac.insert(str);
108 }
109 ac.build();
110 printf("%d\n",ac.solve(m));
111 }
112 return 0;
113 }