Ultraviolet A 1401 remember the word

Trie + dp

Dictionary tree Training Questions in the big white book.

The question is how many small strings a string may contain.

For example

ABCD

4

A

B

CD

AB

ABCD = a + B + CD; ABCD = AB + CD;

Recursion: forward from the last digit, DP [I] = DP [I] + dp [I + Len [x] X is the input sequence and is appended to the node. Yes I ~ The prefix of strlen (s. S [1, 2, 3 ,..., I ,... Len]

When a trie tree is built, the sequence is also attached to the node.

Finally, search for the number of string prefixes and add them together.

#include<cstdio>#include<cstring>#include<string>#include<queue>#include<algorithm>#include<map>#include<stack>#include<iostream>#include<list>#include<set>#include<cmath>#define INF 0x7fffffff#define eps 1e-6#define LL long longusing namespace std;const int MOD=20071027;struct Trie{    int word[4001*100][26];    int ex[4001*100];    int size;    void clear()    {        memset(word[0],0,sizeof(word[0]));        size=1;        ex[0]=0;    }    int insert(char *s,int v)    {        int u=0,c;        for(int i=0; s[i]!='\0'; i++)        {            int c=s[i]-'a';            if(!word[u][c])            {                memset(word[size],0,sizeof(word[size]));                ex[size]=0;                word[u][c]=size++;            }            u=word[u][c];        }        ex[u]=v;    }    int search(char *s,int len,vector<int>& ans)    {        int u=0,c;        for(int i=0; s[i]!='\0'&&i<len; i++)        {            c=s[i]-'a';            if(!word[u][c])return 0;            u=word[u][c];            if(ex[u])                ans.push_back(ex[u]);        }    }} wo;char str[300001];int dp[300001];int n;int le[4001];int main(){    int nn=1;    while(scanf("%s",str)!=EOF)    {        scanf("%d",&n);        wo.clear();        memset(dp,0,sizeof(dp));        char tmp[101];        int len=strlen(str);        for(int i=0; i<n; i++)        {            scanf("%s",tmp);            le[i+1]=strlen(tmp);            wo.insert(tmp,i+1);        }        dp[len]=1;        for(int i=len-1; i>=0; i--)        {            vector<int>ans;            wo.search(str+i,len-i,ans);            for(int j=0; j<ans.size(); j++)                dp[i]=(dp[i]+dp[i+le[ans[j]]])%MOD;        }        printf("Case %d: %d\n",nn++,dp[0]);    }}