Uva11468（AC自动机+概率dp）

坑死我了，其实我还是不太懂为什么一定要补上不存在的边？
但是这题的妙处就是可以由fail跑出所有的后缀来

#include <iostream>
#include <algorithm>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <set>
#include <cmath>
#include <map>
#define LL long long
#define INF 0x3f3f3f3f
#define mod 1000000007
const int maxn = 10000+5;
using namespace std;
char p[1000];
int n,k,l;
double dp[maxn][105];
bool vis[maxn][105];
double pos[1000];
struct Aho{
    struct node{
        int next[63];
        int fail,cnt;
    }state[maxn];
    queue<int> q;
    int size;
    int idx(char ch) {
        if(islower(ch))
            return ch - 'a';
        else if(isupper(ch))
            return ch - 'A' + 26;
        return ch - '0' + 52;
    }
    void init(){
        while(!q.empty()) q.pop();
        for(int i=0; i<maxn; i++){
            memset(state[i].next, 0, sizeof(state[i].next));
            state[i].fail = state[i].cnt = 0;
        }
        size = 1;
    }
    void insert(char *s){
        int n = (int)strlen(s);
        int now = 0;
        for(int i=0; i<n; i++){
            int c = idx(s[i]);
            if(!state[now].next[c]){
                state[now].next[c] = size++;
            }
            now = state[now].next[c];
        }
        state[now].cnt = 1;
    }
    void build(){
        state[0].fail = -1;
        q.push(0);//0是根节点
        while(!q.empty()){
            int u = q.front();
            q.pop();
            for(int i=0; i<62; i++){
                if(state[u].next[i]){
                    if(u == 0) state[state[u].next[i]].fail = 0;
                    else{
                        int v = state[u].fail;//父亲的fail
                        while(v != -1){
                            if(state[v].next[i]){//如果该节点的儿子有这条边
                                state[state[u].next[i]].fail = state[v].next[i];
                                state[state[u].next[i]].cnt |= state[state[v].next[i]].cnt;
                                break;
                            }
                            v = state[v].fail;
                        }
                        if(v == -1)
                            state[state[u].next[i]].fail = 0;
                    }
                    q.push(state[u].next[i]);
                }
                else{//按照蓝书上的话说 是把不存在的fail也补上 导致match时可以不需要不断往上跳
                    if(u == 0) state[u].next[i] = 0;
                    else state[u].next[i] = state[state[u].fail].next[i];
                }
            }
        }
    }
    double match(int u, int l){
        if(!l) return 1.0;
        if(vis[u][l]) return dp[u][l];
        vis[u][l] = true;
        double & ans = dp[u][l];
        ans = 0.0;
        for(int i=0; i<62; i++)
            if(state[state[u].next[i]].cnt == 0) ans += pos[i] * match(state[u].next[i],l-1);
        return ans;
    }
}aho;

int main(){
    int T, kases = 1;
    scanf("%d",&T);
    while(T--){
        aho.init();
        memset(vis, false, sizeof(vis));
        memset(dp, 0, sizeof(dp));
        memset(pos, 0, sizeof(pos));
        scanf("%d",&k);
        for(int i=0; i<k; i++){
            scanf("%s",p);
            aho.insert(p);
        }
        aho.build();
        scanf("%d",&n);
        char ch;
        double ps;
        for(int i=0; i<n; i++){
            getchar();
            scanf("%c %lf",&ch,&ps);
            int c = aho.idx(ch);
            pos[c] = ps;
        }
        scanf("%d",&l);
        printf("Case #%d: %.6lf\n",kases++,aho.match(0,l));
    }
}
/*
2
1
a
2
a 0.5
b 0.5
2

2
ab
ab
2
a 0.2
b 0.8
2
 */