UVa 11468 (AC自动机 概率DP) Substring

将K个模板串构成一个AC自动机，那些能匹配到的单词节点都称之为禁止节点。

然后问题就变成了在Tire树上走L步且不经过禁止节点的概率。

根据全概率公式用记忆化搜索求解。

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

const int maxnode = 500;
const int sigma_size = 64;
int idx[256];

struct AhoCorasickAutomata
{
    int ch[maxnode][sigma_size];
    int match[maxnode];
    int f[maxnode];
    int sz;

    void init() { sz = 1; memset(ch[0], 0, sizeof(ch[0])); }

    void insert(char* s)
    {
        int u = 0, n = strlen(s);
        for(int i = 0; i < n; i++)
        {
            int c = idx[s[i]];
            if(!ch[u][c])
            {
                memset(ch[sz], 0, sizeof(ch[sz]));
                match[sz] = 0;
                ch[u][c] = sz++;
            }
            u = ch[u][c];
        }
        match[u] = 1;
    }

    void getFail()
    {
        queue<int> q;
        f[0] = 0;
        for(int c = 0; c < sigma_size; c++)
        {
            int u = ch[0][c];
            if(u) { f[u] = 0; q.push(u); }
        }
        while(!q.empty())
        {
            int r = q.front(); q.pop();
            for(int c = 0; c < sigma_size; c++)
            {
                int u = ch[r][c];
                if(!u) { ch[r][c] = ch[f[r]][c]; continue; }
                q.push(u);
                int v = f[r];
                while(v && !ch[v][c]) v = f[v];
                f[u] = ch[v][c];
                match[u] |= match[f[u]];
            }
        }
    }
}ac;

int n;
const int maxl = 100 + 10;
char s[30][30];
double prob[sigma_size];

int vis[maxnode][maxl];
double d[maxnode][maxl];

double getProb(int u, int L)
{
    if(L == 0) return 1.0;
    if(vis[u][L]) return d[u][L];
    vis[u][L] = 1;
    double& ans = d[u][L];
    ans = 0;
    for(int c = 0; c < n; c++)
        if(!ac.match[ac.ch[u][c]])
            ans += prob[c] * getProb(ac.ch[u][c], L-1);
    return ans;
}

int main()
{
    //freopen("in.txt", "r", stdin);

    int T;
    scanf("%d", &T);
    for(int kase = 1; kase <= T; kase++)
    {
        int k, L;
        scanf("%d", &k);
        for(int i = 0; i < k; i++) scanf("%s", s[i]);

        scanf("%d", &n);
        for(int i = 0; i < n; i++)
        {
            char s1[9];
            scanf("%s%lf", s1, &prob[i]);
            idx[s1[0]] = i;
        }

        ac.init();
        for(int i = 0; i < k; i++) ac.insert(s[i]);
        ac.getFail();
        scanf("%d", &L);
        memset(vis, 0, sizeof(vis));
        printf("Case #%d: %.6f
", kase, getProb(0, L));
    }

    return 0;
}