POJ2778 AC自动机 + 快速矩阵幂

http://poj.org/problem?id=2778

做法：利用AC自动机建矩阵之后进行N次矩阵乘

关于AC自动机配快速矩阵幂的理解：
1.题目限制10个字符串长度最多为10,那么建出的AC自动机的结点数至多为100

2.任意合法字符串必定通过nxt指针在AC自动机的结点之间转移

3.那么我们只要求出每次结点之间转移的数量，建立一个矩阵，就可以通过快速矩阵幂优化了

4.对于不合法的结点（病毒），将特定的转移次数设定为0即可。

5.注意不合法的结点除了插入的时候字典树上不合法的结点之外，所有fail指针指向不合法结点的及其后缀都是不合法结点，因为fail指针指向结点是该节点的后缀

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double PI = acos(-1.0);
const double eps = 1e-9;
const int maxn = 10010;
const int INF = 0x3f3f3f3f;
const int mod = 100000; 
int M,K;
LL N;
int nxt[maxn][4],fail[maxn],ed[maxn],root,tot;
struct Mat{
    LL a[102][102];
    void init(){Mem(a,0);}
    friend Mat operator *(Mat a,Mat b){
        Mat ans; ans.init();
        for(int i = 0 ; i < tot; i ++){
            for(int j = 0 ; j < tot; j ++){
                for(int k = 0 ; k < tot; k ++){
                    ans.a[i][j] = (ans.a[i][j] + a.a[i][k] * b.a[k][j]) % mod;
                }
            }
        }
        return ans;
    }
    friend Mat operator ^(Mat a,LL t){
        Mat ans = a; t--;
        while(t){
            if(t & 1) ans = ans * a;
            a = a * a;
            t >>= 1;
        }
        return ans;
    }
}base;
int newnode(){
    for(int i = 0 ; i < 4; i ++) nxt[tot][i] = -1;
    ed[tot++] = 0;
    return tot - 1;
}

void init(){
    tot = 0;
    root = newnode();
    base.init();
}
int getid(char c){
    if(c == 'A') return 0;
    if(c == 'C') return 1;
    if(c == 'T') return 2;
    return 3;
}
void insert(char *str){
    int p = root;
    for(int i = 0;str[i]; i ++){
        int id = getid(str[i]);
        if(nxt[p][id] == -1) nxt[p][id] = newnode();
        p = nxt[p][id];
    }
    ed[p] = 1;
}
void Build(){
    queue<int>Q;
    fail[root] = root;
    for(int i = 0 ; i < 4; i ++){
        if(~nxt[root][i]){
            fail[nxt[root][i]] = root;
            Q.push(nxt[root][i]);
        }else{
            nxt[root][i] = root;
        }
    }
    while(!Q.empty()){
        int u = Q.front(); Q.pop();
        for(int i = 0 ; i < 4; i ++){
            if(~nxt[u][i]){
                fail[nxt[u][i]] = nxt[fail[u]][i];
                if(ed[nxt[fail[u]][i]]) ed[nxt[u][i]] = 1;
                Q.push(nxt[u][i]);
            }else{
                nxt[u][i] = nxt[fail[u]][i];
            }
        }
    }
    for(int i = 0 ; i < tot; i ++){
        if(ed[i]) continue;
        for(int j = 0 ; j < 4; j ++){
            if(!ed[nxt[i][j]]) base.a[i][nxt[i][j]]++;
        }
    }    
}
int query(LL q){
    int ans = 0;
    Mat t; t.init();
    t.a[0][0] = 1;
    t = t * (base ^ q);
    for(int i = 0 ; i < tot; i ++){
        for(int j = 0 ; j < tot; j ++) ans = (ans + t.a[i][j]) % mod;
    }    
    return ans;
}
char str[maxn];
int main(){
    Sca(M); Scl(N); init();
    for(int i = 1; i <= M ; i ++){
        scanf("%s",str);
        insert(str);
    }    
    Build();
    Pri(query(N));
    return 0;
}