POJ 2778 DNA Sequence

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

POJ Monthly--2006.03.26

 

解题：真TMD的卡时间，奶奶个熊

#include <iostream>
#include <stdio.h>
#include <cstring>
#include <queue>
using namespace std;
typedef long long LL;
const int mod = 100000;
const int maxn = 80;
int id[256];
struct Matrix{
    int n,m[maxn][maxn];
    void init(int sz){
        n = sz;
        memset(m,0,sizeof m);
    }
    Matrix(int sz){
        init(sz);
    }
    void setOne(){
        for(int i = 0; i < n; ++i) m[i][i] = 1;
    }
    Matrix operator*(const Matrix &rhs){
        Matrix ret(n);
        for(int k = 0; k < n; ++k)
            for(int i = 0; i < n; ++i)
                for(int j = 0; j < n; ++j)
                    ret.m[i][j] = (ret.m[i][j] + (LL)m[i][k]*rhs.m[k][j]%mod)%mod;
        return ret;
    }
    void out(){
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < n; ++j)
                cout<<m[i][j]<<" ";
            cout<<endl;
        }
    }
};
void quickPow(Matrix &base,Matrix &ret,int index){
    ret.setOne();
    while(index){
        if(index&1) ret = ret*base;
        index >>= 1;
        base = base*base;
    }
}
struct Trie{
    int ch[maxn][4],fail[maxn],cnt[maxn],tot;
    void init(){
        tot = 0;
        newnode();
    }
    int newnode(){
        memset(ch[tot],0,sizeof ch[tot]);
        fail[tot] = cnt[tot] = 0;
        return tot++;
    }
    void insert(char *str,int root = 0){
        for(int i = 0; str[i]; ++i){
            int &x = ch[root][id[str[i]]];
            if(!x) x = newnode();
            root = x;
        }
        ++cnt[root];
    }
    void build(int root = 0){
        queue<int>q;
        for(int i = 0; i < 4; ++i)
            if(ch[root][i]) q.push(ch[root][i]);
        while(!q.empty()){
            root = q.front();
            q.pop();
            cnt[root] += cnt[fail[root]];
            for(int i = 0; i < 4; ++i){
                int &x = ch[root][i];
                if(x){
                    fail[x] = ch[fail[root]][i];
                    q.push(x);
                }else x = ch[fail[root]][i];
            }
        }
    }
    int solve(int m){
        Matrix a(tot),b(tot);
        for(int i = 0; i < tot; ++i){
            if(cnt[i]) continue;
            for(int j = 0; j < 4; ++j){
                int x = ch[i][j];
                if(cnt[x]) continue;
                ++a.m[i][x];
            }
        }
        quickPow(a,b,m);
        int ret = 0;
        for(int i = 0; i < tot; ++i)
            ret = (ret + b.m[0][i])%mod;
        return ret;
    }
}ac;
int main(){
    for(int i = 0; i < 4; ++i) id["ATCG"[i]] = i;
    char str[maxn];
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        ac.init();
        for(int i = 0; i < n; ++i){
            scanf("%s",str);
            ac.insert(str);
        }
        ac.build();
        printf("%d\n",ac.solve(m));
    }
    return 0;
}