BZOJ 1444 有趣的游戏(AC自动机+矩阵快速幂)

真的是很有趣的游戏。。。

对每个单词构建好AC自动机后，由于单词都是相同长度的且不同，所以不会出现互相为子串的形式。

那么我们对AC自动机上的节点构建转移矩阵。对于每个单词末尾的节点。该节点的出边仅仅与自己相连且概率为1.

表示如果已经出现了该单词游戏就结束了。答案是收敛的，我们对这个矩阵迭代个2^50次应该就可以求出近似的答案了。

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi 3.1415926535
# define eps 1e-4
# define MOD 1000000007
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
inline int Scan() {
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=105;
//Code begin...

struct Matrix{double matrix[N][N];}sa;
int trie[N][12], top, fail[N], m, pos[N];
char s[N];
double P[N];

void init(){top=1; mem(trie[0],0);}
void ins(char *s, int i){
    int rt, nxt;
    for (rt=0; *s; rt=nxt, ++s){
        nxt=trie[rt][*s-'A'];
        if (!nxt) mem(trie[top],0), trie[rt][*s-'A']=nxt=top++;
    }
    ++trie[rt][10]; pos[i]=rt;
}
void makefail(){
    int u, v, bg, ed;
    static int q[N];
    fail[0]=bg=ed=0;
    FO(i,0,m) if ((v=trie[0][i])) fail[q[ed++]=v]=0;
    while (bg<ed){
        u=q[bg++];
        FO(i,0,m) {
            if ((v=trie[u][i])) fail[q[ed++]=v]=trie[fail[u]][i];
            else trie[u][i]=trie[fail[u]][i];
        }
    }
}
Matrix Mul(Matrix a, Matrix b) //矩阵乘法(%MOD)
{
    Matrix c;
    FO(i,0,top) FO(j,0,top) {
          c.matrix[i][j]=0;
          FO(l,0,top) c.matrix[i][j]+=a.matrix[i][l]*b.matrix[l][j];
    }
    return c;
}
int main ()
{
    int n, l, nxt;
    double x, y;
    scanf("%d%d%d",&n,&l,&m); init();
    FO(i,0,m) scanf("%lf%lf",&x,&y), P[i]=x/y;
    FO(i,0,n) scanf("%s",s), ins(s,i);
    makefail();
    FO(i,0,top) {
        if (trie[i][10]) {sa.matrix[i][i]=1; continue;}
        FO(j,0,m) nxt=trie[i][j], sa.matrix[i][nxt]+=P[j];
    }
    FOR(i,1,50) sa=Mul(sa,sa);
    FO(i,0,n) printf("%.2f\n",sa.matrix[0][pos[i]]);
    return 0;
}