LA 3942 背单词（dp+Trie）

分析：d[i]=sum{ d(i+len(x) } [i,L]的种类数

可以正向枚举，也可以逆向枚举，Trie+dp的结合题，第一次见，其实dp的思维还是比较简单的

<pre class="cpp" name="code">#include <cstdio>#include <cstring>#include <string>#include <iostream>#include <algorithm>using namespace std;const int mod=20071027;struct TrieNode{bool vis;struct TrieNode *next[26];};typedef TrieNode Trie;Trie *root;char str[300006];long long dp[300006];Trie *CreateNode(){Trie *p=(Trie*)malloc(sizeof(Trie));p->vis=false;for (int i=0;i<26;i++)p->next[i]=NULL;return p;}void Insert(string str){Trie *p=root;int len=str.size();for (int i=0;i<len;i++){int id=str[i]-'a';if (p->next[id] == NULL){p->next[id]=CreateNode();p=p->next[id];}else p=p->next[id];}p->vis=true;}void Search(int ii,int len){Trie *p=root;dp[ii]=0;for (int i=ii;i<len;i++){int id=str[i]-'a';p=p->next[id];if (p == NULL) break;if (p->vis == true) dp[ii]=(dp[ii] + dp[i+1]) % mod;}}int main(){int j=0;while (~scanf("%s",str)){root=CreateNode();int len=strlen(str);int n;scanf("%d",&n);string s;for (int i=1;i<=n;i++){cin>>s;Insert(s);}dp[len]=1;//精髓for (int i=len-1;i>=0;i--){Search(i,len);}printf("Case %d: %d\n",++j,dp[0]);}}