bzoj 1444 AC自动机 + 矩阵乘法 | 高斯消元

恶补了一下AC自动机，花了一天时间终于全部搞明白了。

思路：将每个人的串加入AC自动机，在AC自动机生成的状态图上建边，注意单词末尾的节点只能转移到自己概率为1，

然后将矩阵自乘几十次后误差就很小了， 或者可以高斯消元搞出精确解。

#include<bits/stdc++.h>
#define LL long long
#define ll long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>
#define y1 skldjfskldjg
#define y2 skldfjsklejg

using namespace std;

const int N =  + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;

int n, l, m, pos[];
double pro[];
char s[];

struct Matrix {
    int r, c;
    double a[][];
    Matrix(int r = , int c = ) {
        this->r = r;
        this->c = c;
        memset(a, , sizeof(a));
    }

    Matrix operator * (const Matrix &B) const {
        Matrix C(r, c);
        for(int i = ; i < r; i++)
            for(int j = ; j < c; j++)
                for(int k = ; k < r; k++)
                    C.a[i][j] += a[i][k] * B.a[k][j];
        return C;
    }
};

struct Ac {
    int val[N], ch[N][], f[N], last[N], cnt, SZ;

    void init(int SZ = ) {
        cnt = ; this->SZ = SZ;
        for(int c = ; c < SZ; c++) ch[][c] = ;
    }

    int getId(char c) {
        return c - 'A';
    }

    int newNode() {
        cnt++;
        memset(ch[cnt], , sizeof(ch[cnt]));
        val[cnt] = f[cnt] = last[cnt] = ;
        return cnt;
    }

    void add(char *s, int &pos) {
        int u = ;
        for(int i = ; s[i]; i++) {
            int c = getId(s[i]);
            if(!ch[u][c]) ch[u][c] = newNode();
            u = ch[u][c];
        }
        val[u]++;
        pos = u;
    }

    void build() {
        queue<int> que;
        f[] = ;
        for(int c = ; c < SZ; c++) {
            if(!ch[][c]) continue;
            f[ch[][c]] = last[ch[][c]] = ;
            que.push(ch[][c]);
        }
        while(!que.empty()) {
            int u = que.front(); que.pop();
            for(int c = ; c < SZ; c++) {
                int v = ch[u][c];
                if(!v) {
                    ch[u][c] = ch[f[u]][c];
                    continue;
                } else {
                    que.push(v);
                    f[v] = ch[f[u]][c];
                    last[v] = val[f[v]] ? f[v] : last[f[v]];
                }
            }
        }
    }

    void buildMatrix(Matrix &A) {
        for(int u = ; u <= cnt; u++) {
            if(val[u]) A.a[u][u] = ;
            else {
                for(int c = ; c < m; c++) {
                    int v = ch[u][c];
                    A.a[u][v] += pro[c];
                }
            }
        }
    }
} ac;

int main() {
    scanf("%d%d%d", &n, &l, &m);
    for(int i = ; i < m; i++) {
        double p, q;
        scanf("%lf%lf", &p, &q);
        pro[i] = p / q;
    }

    ac.init(m);

    for(int i = ; i <= n; i++) {
        scanf("%s", s);
        ac.add(s, pos[i]);
    }

    ac.build();
    Matrix A(ac.cnt + , ac.cnt + );
    ac.buildMatrix(A);

    for(int i = ; i <= ; i++)
        A = A * A;

    for(int i = ; i <= n; i++) printf("%.2f\n", A.a[][pos[i]]);
    return ;
}

/*
*/