POJ2778 AC自动机 + 快速矩阵幂

傷城~ 2023-10-11 13:28 35阅读 0赞

> 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;
    }

转载于:https://www.cnblogs.com/Hugh-Locke/p/11437136.html