Ultraviolet A 1401-Remember the Word // TRie

This question is mainly implemented in recursion.

Set a word W, which is composed of (w1, w2, ·, wn) letters to determine the combination of words.

Given any suffix (wi, wi + 1,..., wn), you must find its prefix to determine the number of combinations. Set the suffix (wi, wi + 1,..., wn) to x, find the number of prefixes in the suffix (the prefix here refers to the word ). Set d (I) to indicate the number of combination methods for the Suffix from I to n, set x to the prefix of the suffix, and the remaining part of the string is y. Then d (I) = d (len (x) + I ).

Why? A prefix corresponds to at least one combination, but the group prefix is not necessarily consistent. You need to determine whether y is a word. Therefore, you need to determine the number of combinations of d (len (x) + I. Because a suffix has multiple prefixes, We Need To sum all prefixes. There is a recursive relationship below:

D (I) = sum {d (len (x) + I )}

Therefore, you can use the Trie tree to quickly find the prefix.

The following code is used:

#include<stdio.h>#include<string.h>#define maxn 500010int ch[maxn][26];int val[maxn],cnt,n;int d[maxn];char str[maxn],s[110];void initi(){    cnt = 1;    memset(ch[0],0,sizeof(ch[0]));    memset(d,0,sizeof(d));}void insert_Trie(char *s){    int u = 0;    while(*s)    {        int id = *s - 'a';        if(ch[u][id] == 0)        {            memset(ch[cnt],0,sizeof(ch[cnt]));            val[cnt] = 0;            ch[u][id] = cnt++;        }        u = ch[u][id];        s++;    }    val[u] = -1;}int main(){    int c = 0;    while(scanf("%s",str) != EOF)    {        c++;        initi();        scanf("%d",&n);        for(int i = 0; i < n; i++)        {            scanf("%s",s);            insert_Trie(s);        }        int len = strlen(str);        d[len] = 1;        for(int i = len-1; i >= 0; i--)        {            int j,k;            int u = 0;            for(j = 0,k = i; k < len; j++,k++)            {                int id = str[k] - 'a';                if(!ch[u][id]) break;                else if(val[ch[u][id]] == -1)                {                    d[i] += d[i+j+1];                    if(d[i] >= 20071027) d[i] %= 20071027;                }                u = ch[u][id];            }        }        printf("Case %d: ",c);        printf("%d\n",d[0]%20071027);    }    return 0;}

Note:

1. query operations are not required when prefix search is performed.

2. The result is very large, and int will pop up. You need to handle it because it is mod.