解题思路：
判断给出的单词是否恰好由另外两个单词组成，用栈保存每个子字符串的节点，，从这个节点出发判断剩下的字符串是否在字典树中即可。
#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <map>#include <stack>#include <set>using namespace std;const int MAXN = 50000 + 10;typedef struct Trie_Node{bool isword;struct Trie_Node *next[26];}Trie;void insert(Trie *root, char *word){Trie *p = root;int i = 0;while(word[i] != '\0'){if(p->next[word[i]-'a'] == NULL){Trie *temp = new Trie;temp->isword = false;for(int j=0;j<26;j++)temp->next[j] = NULL;p->next[word[i]-'a'] = temp;}p = p->next[word[i]-'a'];i++;}p->isword = true;}bool Find(Trie *root, char *word){Trie *p = root;int i = 0;stack<int> s;while(!s.empty()) s.pop();while(word[i] != '\0'){if(p->next[word[i]-'a'] == NULL)return 0;p = p->next[word[i]-'a'];if(p->isword && word[i]) s.push(i+1);i++;}while(!s.empty()){bool ok = true;i = s.top(); s.pop();p = root;while(word[i]){if(p->next[word[i]-'a'] == NULL){ok = false;break;}p = p->next[word[i]-'a'];i++;}if(ok && p->isword)return 1;}return 0;}char s[MAXN][27];int main(){int n = 0;Trie *root = new Trie;root->isword = false;for(int j=0;j<26;j++)root->next[j] = NULL;while(scanf("%s", s[n]) != EOF){insert(root, s[n]);n++;}for(int i=0;i<n;i++){if(Find(root, s[i]))printf("%s\n", s[i]);}return 0;}
记忆像是倒在手心里的水，不论是摊平还是握紧，
