UVa 11468

方法：ac自动机 dp

方法比较明显，dp[cur][left] 表示在cur结点，能走left步的期望。在ac自动机上记忆化搜索即可。注意，判断下一个结点nxt是否可行，要看val[nxt] 和 last[nxt]。所以在get fail 的时候，也可省去last，直接 val[nxt] |= val[f[nxt]]。（val[i] = 1 代表 i结点为某个禁止串的终点，＝0则不是）。

code:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#include <fstream>
#include <cassert>
#include <unordered_map>
#include <unordered_set>
#include <cmath>
#include <sstream>
#include <time.h>
#include <complex>
#include <iomanip>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a))
#define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a))
#define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a))
#define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a))
#define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
#define FOREACH(a,b) for (auto &(a) : (b))
#define rep(i,n) FOR(i,0,n)
#define repn(i,n) FORN(i,1,n)
#define drep(i,n) DFOR(i,n-1,0)
#define drepn(i,n) DFOR(i,n,1)
#define MAX(a,b) a = Max(a,b)
#define MIN(a,b) a = Min(a,b)
#define SQR(x) ((LL)(x) * (x))
#define Reset(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(v) v.begin(),v.end()
#define ALLA(arr,sz) arr,arr+sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(all(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr,sz) sort(ALLA(arr,sz))
#define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
#define PERMUTE next_permutation
#define TC(t) while(t--)
#define forever for(;;)
#define PINF 1000000000000
#define newline '
'

#define test if(1)if(0)cerr
using namespace std;
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;
typedef pair<double,double> dd;
typedef pair<char,char> cc;
typedef vector<ii> vii;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> l4;
const double pi = acos(-1.0);

const int MAXNODE = 20*20+5;
const int SIGMA_SIZE = 130;
const int MAXL = 105;
inline int idx(char c)
{
    return c;
}

struct AC
{
    int g[MAXNODE][SIGMA_SIZE], val[MAXNODE], last[MAXNODE], f[MAXNODE];
    int sz;
    int newnode()
    {
        Reset(g[sz], 0);
        val[sz] = last[sz] = 0;
        return sz++;
    }
    void init()
    {
        sz = 0; newnode();
    }
    void insert(const string &str, int v)
    {
        int u = 0, n = str.length();
        rep(i, n)
        {
            int c = idx(str[i]);
            if (!g[u][c]) g[u][c] = newnode();
            u = g[u][c];
        }
        val[u] = v;
    }
    void get_fail()
    {
        queue<int> q;   f[0] = 0;
        rep(i, SIGMA_SIZE)
        {
            int u = g[0][i];
            if (u)
            {
                f[u] = last[u] = 0; q.push(u);
            }
        }
        while (!q.empty())
        {
            int r = q.front();  q.pop();
            rep(c, SIGMA_SIZE)
            {
                int u = g[r][c];
                if (!u)
                {
                    g[r][c] = g[f[r]][c]; continue;
                }
                q.push(u);
                int v = f[r];
                while (v && !g[v][c]) v = f[v];
                f[u] = g[v][c];
                last[u] = val[f[u]]?f[u]:last[f[u]];
            }
        }
    }
    void valid(int j)
    {
        if (j)
        {
            valid(last[j]);
        }
    }
    void find(const string &str)
    {
        int n = str.length(), j = 0;
        rep(i, n)
        {
            int c = idx(str[i]);
            j = g[j][c];
            if (val[j]) valid(j);
            else if (last[j]) valid(last[j]);
        }
    }
};
AC ac;

map<int, double> p;
string str;
int k, l, N;
double d[MAXNODE][MAXL];
bool vis[MAXNODE][MAXL];
double dp(int cur, int left)
{
    if (!left) return 1.0;
    if (!vis[cur][left])
    {
        vis[cur][left] = true;
        double &ans = d[cur][left];
        ans = 0;
        for (auto &pr : p) if (!ac.val[ac.g[cur][pr.first]] && !ac.last[ac.g[cur][pr.first]])
        {
            ans += pr.second*dp(ac.g[cur][pr.first], left-1);
        }
    }
    return d[cur][left];
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;  cin >> T;
    repn(kase, T)
    {
        ac.init();
        p.clear();
        cin >> k;
        rep(i, k)
        {
            string line;    cin >> line;
            ac.insert(line, 1);
        }
        ac.get_fail();
        cin >> N;
        rep(i, N)
        {
            char c; cin >> c;
            double temp;    cin >> temp;
            p[idx(c)] = temp;
        }
        cin >> l;
        Reset(vis, 0);
        printf("Case #%d: %.6lf
", kase, dp(0, l));
    }
}