La3942 remember the word (trie + dp)

Simple application of the trie graph. The key to this question is to come up with a recursive formula. D (I) indicates the string starting with character I, d (I) = sum {d (I + Len (x)}, X is s [I... l] prefix. Then, all words that can be broken down into a trie tree and then run the parent string on it. D [0] is the total number of solutions.

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define mod 20071027#define M 400005using namespace std;int n,top,len;int tree[M][27];char S[M];char p[105];int val[M];int d[M];void init(){    top=1;    memset(tree,0,sizeof(tree));    memset(d,0,sizeof(d));}int idx(char c){    return c-'a';}void insert(char *s){    int Len=strlen(s);    int u=0;    for(int i=0;i<Len;i++)    {        int c=idx(s[i]);        if(!tree[u][c])        {            val[top]=0;            tree[u][c]=top++;        }        u=tree[u][c];    }    val[u]=1;}int query(char *s,int start){    int count=0;    int u=0;    for(int i=start;i<len;i++)    {        int c=idx(s[i]);        u=tree[u][c];        if(!u) return count;        if(val[u])        {            count+=d[i+1];            count%=mod;        }    }    return count;}int main(){    int t=1;    //freopen("d:\\test.txt","r",stdin);    while(scanf("%s",S)!=EOF)    {        scanf("%d",&n);        init();        for(int i=0;i<n;i++)        {            scanf("%s",p);            insert(p);        }        len=strlen(S);        d[len]=1;        for(int i=1;i<=len;i++)        {            d[len-i]=query(S,len-i);        }        cout<<"Case "<<t++<<": "<<d[0]<<endl;    }    return 0;}

La3942 remember the word (trie + dp)