bzoj1444[Jsoi2009]有趣的游戏[AC自动机]

题面

bzoj

我要向师父学习善待每一只数据结构

考虑成环，那么高斯消元

然鹅这道题太小了 所以直接转移矩阵自乘就好啦

终点不向外连边 有一条向自己的，概率为一的自环来作为结尾

对于其他店 若有边\((u -> v) = p\) 那么mat[u][v] += p

#include <cmath>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <complex>
#include <ctime>
#include <vector>
#include <queue>
#include <bitset>
#define mp(x, y) make_pair(x, y)
using namespace std;
const int N = 15;
const int M = 105;
int n, m, len, sz, en[N];
double p[N];
struct Matrix{
	double w[M][M];
	void clear(){
		for(int i = 0; i <= sz; ++i)
		     for(int j = 0; j <= sz; ++j)
		         w[i][j] = 0;
	}
	void print(){
		printf("--------------------\n");
		for(int i = 0; i <= sz; ++i){
		     for(int j = 0; j <= sz; ++j)
		         printf("%.2lf ", w[i][j]);
			printf("\n");
		}
		printf("--------------------\n");
	}
	friend Matrix operator *(const Matrix x, const Matrix y){
		Matrix z; z.clear();
		for(int i = 0; i <= sz; ++i)
		    for(int j = 0; j <= sz; ++j)
		        for(int k = 0; k <= sz; ++k)
		            z.w[i][j] += x.w[i][k] * y.w[k][j];
	//	z.print();
	    return z;
	}
}mat, res;
struct AC{
	int ch[M][N], f[M];
	bool flag[M];
	queue<int> que;
	void ins(char* str, int id){
		int now = 0;
		for(int i = 1, cc; i <= len; ++i){
			cc = str[i] - 'A';
			if(!ch[now][cc]) ch[now][cc] = ++sz;
			now = ch[now][cc];
		}
		en[id] = now, flag[now] = 1;
	}
	void build(){
		int now = 0;
		for(int i = 0; i < m; ++i) if(ch[0][i]) que.push(ch[0][i]);
		while(!que.empty()){
			int fro = que.front(); que.pop();
			for(int i = 0; i < m; ++i){
				if(ch[fro][i]) f[ch[fro][i]] = ch[f[fro]][i], que.push(ch[fro][i]);//!!
				else ch[fro][i] = ch[f[fro]][i];
			}
		}
		mat.clear();
		for(int i = 0; i <= sz; ++i){
			if(flag[i]){
				mat.w[i][i] = 1; continue;
			}
		    for(int j = 0; j < m; ++j){
		    	mat.w[i][ch[i][j]] += p[j];
		    }
		}
	}
}ac;

int main(){
    scanf("%d%d%d", &n, &len, &m);
    for(int i = 0; i < m; ++i){
    	double x, y; scanf("%lf%lf", &x, &y);
    	p[i] = x / y;
    }
    char str[N];
    for(int i = 1; i <= n; ++i){
    	scanf("%s", str + 1);
    	ac.ins(str, i);
    }
    ac.build();
    for(int i = 1; i <= 100; ++i) mat = mat * mat;
    //转移矩阵自乘 得到来自0的解
	for(int i = 1; i <= n; ++i) printf("%.2lf\n", mat.w[0][en[i]]);
    return 0;
}