hdu 3689 Infinite monkey theorem

去年杭州区域赛的一道题目，比较简单的AC自动机

#include <stdio.h>#include <string.h>const int NODE = 1000;const int CH = 26;int tree[NODE][CH], cnt;int fail[NODE], word[NODE], Que[NODE];int num[128];char str[20];double gl[128];double dp[1010][20];int mark;int n, m;double sum;int Ins(char *a, int val){    int p = 0;    for(; *a; a++)    {        int c = num[*a];        if(!tree[p][c])        {            memset(tree[cnt], 0, CH * sizeof(int));            word[cnt] = 0;            tree[p][c] = cnt++;        }        p = tree[p][c];    }    word[p] += val;    return p;}void AC(){    int *s = Que, *e = Que;    for(int i = 0; i < CH; i++)        if(tree[0][i])        {            fail[tree[0][i]] = 0;            *e++ = tree[0][i];        }    while(s != e)    {        int p = *s++;        for(int i = 0; i < CH; i++)        {            if(tree[p][i])            {                int v = tree[p][i];                *e++ = v;                fail[v] = tree[fail[p]][i];            }            else            {                tree[p][i] = tree[fail[p]][i];            }        }    }}void solve(){    memset(dp, 0, sizeof(dp));    sum = 0;    dp[0][0] = 1;    for(int i = 0; i < m; i++)    {        for(int j = 0; j < cnt; j++)        if(i >= j)        {            for(int k = 0; k < CH; k++)            {                int v = tree[j][k];                dp[i + 1][v] += dp[i][j] * gl[k];            }        }        if(dp[i+1][mark] >= 0)        {            sum += dp[i+1][mark];            dp[i+1][mark] = 0;        }    }    /*for(int i = 1; i < m; i++)    {        for(int j = 0; j < cnt; j++)        printf("dp[%d][%d] = %0.4lf   ", i, j, dp[i][j]);        printf("\n");    }*/    printf("%0.2lf%%\n", sum * 100);}int main(){    for(int i = 0; i < 26; i++)  num[i + 'a'] = i;    while(scanf("%d%d", &n, &m) != EOF)    {        if(n == 0 && m == 0)  break;        fail[0] = 0;        cnt = 1;        memset(tree[0], 0, CH * sizeof(int));        memset(gl, 0, sizeof(gl));        for(int i = 1; i <= n; i++)        {            double x;            scanf("%s%lf", str, &x);            gl[str[0] - 'a'] = x;        }        scanf("%s", str);        mark = Ins(str, 1);        AC();        solve();    }    return 0;}