Hat’s Words

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链接：http://acm.hust.edu.cn/vjudge/problem/18360/origin

题目：A hat’s word is a word in the dictionary that is the concatenation of exactly two other words in the dictionary.
You are to find all the hat’s words in a dictionary.

题意：给出一个单词表，把所有可以用另外两个单词链接起来表示的单词打印出来。

分析：多维互相匹配，字典树的特征。存储用模板就好，查询的时候要下点功夫。对于这道题，某个单词可能不仅仅能由两个单词组成，比如。。。瞎编一个ab，abcd，cd，c，d。abcd可以是两个也可以是三个单词组成。解决办法是查找的时候，把每个分割点都记录下来，如果分割点到结尾可以组成一个单词，那么就可以被两个单词组成。问题得解。

题解：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <vector>#include <map>#include <string>#include <cstring>#include <functional>#include <cmath>#include <cctype>#include <cfloat>#include <climits>#include <complex>#include <deque>#include <list>#include <set>#include <utility>#define rt return 0#define fr freopen("in.txt","r",stdin)#define fw freopen("out.txt","w",stdout)#define ll long long#define detie ios_base::sync_with_stdio(false);cin.tie(false)using namespace std;struct Node{Node() :is(false), count(0) { memset(next, 0, sizeof next); }bool is;int count;int next[26];}no[100000];string s[50010];int root = 0;int rl = 0;void insert(string &s){int len = s.size();int p = root;for (int i = 0; i < len; i++){int id = s[i] - 'a';if (!(no[p].next[id]))no[p].next[id] = ++rl;p = no[p].next[id];}no[p].is = true;}bool find(string &s){int len = s.size();int sta[1000];int p = root;int sl = 0;for (int i = 0; i < len; i++){int id = s[i] - 'a';p = no[p].next[id];if (no[p].is)sta[sl++] = i+1;}while (sl){bool flag = true;int i = sta[--sl];p = root;while (i<len){int id = s[i++] - 'a';if (!(no[p].next[id])){flag = false;break;}p = no[p].next[id];}if (flag&&no[p].is)return true;}return false;}int main(){//fr;detie;int i = 0;//root = new Node;while (cin>>s[i]){insert(s[i]);i++;}int j = 0;while (j<i){if (find(s[j]))cout << s[j] << endl;j++;}rt;}

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