Substring UVA

题意：

给出一些字符和各自对应的选择概率，随机选择L次后得到一个长度为L的随机字符串S。

给出K个模板串，计算S不包含任何一个模板串的概率

dp【i】【j】表示走到AC自动机 i 这个节点 还需要走 j 步的概率。

表示不会概率DP ，看网上题解写的。

通过记忆化搜索去写。

 注意一点字符有大小字母和数字

#include <set>
#include <map>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>

#define  pi    acos(-1.0)
#define  eps   1e-9
#define  fi    first
#define  se    second
#define  rtl   rt<<1
#define  rtr   rt<<1|1
#define  bug                printf("******
")
#define  mem(a, b)          memset(a,b,sizeof(a))
#define  name2str(x)        #x
#define  fuck(x)            cout<<#x" = "<<x<<endl
#define  sfi(a)             scanf("%d", &a)
#define  sffi(a, b)         scanf("%d %d", &a, &b)
#define  sfffi(a, b, c)     scanf("%d %d %d", &a, &b, &c)
#define  sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
#define  sfL(a)             scanf("%lld", &a)
#define  sffL(a, b)         scanf("%lld %lld", &a, &b)
#define  sfffL(a, b, c)     scanf("%lld %lld %lld", &a, &b, &c)
#define  sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
#define  sfs(a)             scanf("%s", a)
#define  sffs(a, b)         scanf("%s %s", a, b)
#define  sfffs(a, b, c)     scanf("%s %s %s", a, b, c)
#define  sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
#define  FIN                freopen("../in.txt","r",stdin)
#define  gcd(a, b)          __gcd(a,b)
#define  lowbit(x)          x&-x
#define  IO                 iOS::sync_with_stdio(false)


using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const ULL seed = 13331;
const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int maxn = 1e6 + 7;
const int maxm = 8e6 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int T, n, m, L, id[70], vis[410][110];
char buf[410];
double dp[410][110], pro[70];

int get_num(char ch) {
    if (ch >= 'a' && ch <= 'z') return ch - 'a';
    if (ch >= 'A' && ch <= 'Z') return ch - 'A' + 26;
    if (ch >= '0' && ch <= '9') return ch - '0' + 52;
}

struct Aho_Corasick {
    int next[410][66], fail[410], End[410];
    int root, cnt;

    int newnode() {
        for (int i = 0; i < 66; i++) next[cnt][i] = -1;
        End[cnt++] = 0;
        return cnt - 1;
    }

    void init() {
        cnt = 0;
        root = newnode();
    }

    void insert(char buf[]) {
        int len = strlen(buf);
        int now = root;
        for (int i = 0; i < len; i++) {
            if (next[now][get_num(buf[i])] == -1) next[now][get_num(buf[i])] = newnode();
            now = next[now][get_num(buf[i])];
        }
        End[now] = 1;
    }

    void build() {
        queue<int> Q;
        fail[root] = root;
        for (int i = 0; i < 66; i++)
            if (next[root][i] == -1) next[root][i] = root;
            else {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while (!Q.empty()) {
            int now = Q.front();
            Q.pop();
            End[now] |= End[fail[now]];
            for (int i = 0; i < 66; i++)
                if (next[now][i] == -1) next[now][i] = next[fail[now]][i];
                else {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }

    double solve(int pos, int res) {
        if (!res) return 1.0;
        if (vis[pos][res]) return dp[pos][res];
        vis[pos][res] = 1;
        double &ret = dp[pos][res];
        ret = 0;
        for (int i = 0; i < m; ++i) {
            int idx = id[i];
            if (!End[next[pos][idx]]) ret += pro[i] * solve(next[pos][idx], res - 1);
        }
        return ret;
    }

    void debug() {
        for (int i = 0; i < cnt; i++) {
            printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
            for (int j = 0; j < 26; j++) printf("%2d", next[i][j]);
            printf("]
");
        }
    }
} ac;

int main() {
  //  FIN;
    int cas = 1;
    sfi(T);
    while (T--) {
        sfi(n);
        ac.init();
        for (int i = 0; i < n; ++i) {
            sfs(buf);
            ac.insert(buf);
        }
        ac.build();
        sfi(m);
        for (int i = 0; i < m; ++i) {
            scanf("%s%lf", buf, &pro[i]);
            id[i] = get_num(buf[0]);
        }
        sfi(L);
        mem(vis, 0);
        printf("Case #%d: %.6f
", cas++, ac.solve(0, L));
    }
    return 0;
}