poj 2778 ac自动机dp矩阵快速幂

最新推荐文章于 2021-06-13 22:17:32 发布

原创最新推荐文章于 2021-06-13 22:17:32 发布 · 688 阅读

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CC 4.0 BY-SA版权

本文介绍了一种结合矩阵快速幂与AC自动机的方法来高效解决特定问题。通过矩阵压缩和快速幂运算，显著提高了算法效率，从766ms降低到32ms。此外，还详细展示了AC自动机的构建、失败指针设置及如何生成矩阵。

题目链接

参考题解

程序未对矩阵进行压缩时间是766ms，进行压缩后是32ms

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include<queue>
using namespace std;
typedef long long LL;
const int SN = 4;
const int MN = 105;
int N = MN;
const int M = 100000;
struct Matrix{
    int v[MN][MN];
    Matrix(){memset(v,0,sizeof(v));}
    Matrix(int){
        Matrix();
        for(int i = 0; i < N; ++i)v[i][i] = 1;
    }
}a;
//struct queue{
//    int top,rear;
//    int a[N];
//    queue(){top = rear = 0;}
//    int front(){return a[top++];}
//    bool empty(){return top == rear;}
//    void push(int k){a[rear++] = k;}
//};
Matrix operator *(const Matrix &a,const Matrix &b){
            Matrix c;
            long long v;
            for(int i = 0; i < N; ++i)
                for(int j = 0;j < N; ++j){
                    v = 0;
                    for(int k = 0; k < N; ++k)
                        v += (LL)a.v[i][k]*b.v[k][j];
                   if(v >= M)v %= M;
                   c.v[i][j] = (int) v;
                }
            return c;
}
//Matrix operator % (Matrix a,const int m){
//            for(int i = 0; i < N; ++i)
//                for(int j = 0;j < N; ++j)
//                    a.v[i][j] %= M;
//            return a;
//}
//Matrix operator ^(Matrix a,int b){
//            if(b == 1) return a;
//            Matrix c = (a^(b>>1));
//            if(b&1) return c*c*a;
//            return c*c;
//}
Matrix operator^(Matrix a,long long b){
        Matrix c(0);
        while(b){
                if(b&1) c = c*a;
                a = a*a;
                b >>= 1;
        }
        return c;
}

long long getsumoffirstline(Matrix a){
        long long sum = 0;
        for(int i = 0;i < N; ++i) sum += a.v[0][i];
        return sum % M;
}
class AAM{
public:
    int ch[MN][SN],tag[MN],fail[MN];
    int root,cnt;
    int id['Z'];
    void init(){
        cnt = 0,root = cnt++;
        id['A'] = 0;
        id['T'] = 1;
        id['G'] = 2;
        id['C'] = 3;
        memset(ch[0],0,sizeof(ch[0]));
        memset(tag,0,sizeof(tag));
        memset(fail,0,sizeof(fail));
    }
    void insert(char *s){
        int p = root;
        while(*s){
                int  index = id[*s];
                if(tag[p]) return;//小小的优化
                if(!ch[p][index]){
                    ch[p][index] = cnt;

                    memset(ch[cnt],0,sizeof(ch[cnt]));
                    ++cnt;
                }
                p = ch[p][index];
                s++;
        }
        tag[p] = 1;
    }
    void getfail(){
            queue<int> Q;
            for(int i = 0; i < SN; ++i){//两个数组都有清零不用再赋0给他了
                 if(ch[0][i]){
                        Q.push(ch[0][i]);
                 }
            }
            while(!Q.empty()){
                int u = Q.front();Q.pop();
                if(tag[fail[u]]) tag[u] = 1;
                for(int i = 0;i < SN; ++i){
                    int &v = ch[u][i];
                    if(v){
                          fail[v] = ch[fail[u]][i];
                          Q.push(v);
                    }else v = ch[fail[u]][i];
                }
            }
    }
    void getMatrix(){
            Matrix c;
            int cnt1,cnt2;

            for(int i = 0; i < cnt; ++i)
                for(int j = 0; j < SN; ++j)
                    if(!tag[i] && !tag[ch[i][j]]){
                        a.v[i][ch[i][j]] ++;
                    }


            for(int i = 0; i < cnt; ++i){   //压缩行列
                cnt1 = 0;
                for(int j = 0; j < cnt; ++j){
                        if(!tag[j]){
                            a.v[i][cnt1++] = a.v[i][j];
                        }
                }

            }

                for(int i = 0; i < cnt; ++i){
                    cnt1 = 0;
                    for(int j = 0; j < cnt; ++j){
                        if(!tag[j]){
                            a.v[cnt1++][i] = a.v[j][i];
                        }
                    }
                }

        N = cnt1;
    }
};
char s[15];
long long n,m;
AAM T;
int main()
{
    T.init();
    scanf("%lld%lld",&m,&n);
    while(m--){
        scanf("%s",s);
        T.insert(s);
    }
    T.getfail();
    T.getMatrix();
    printf("%lld\n",getsumoffirstline(a^n));
    return 0;
}