HDU 1247 Hat’s Words

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11887    Accepted Submission(s): 4235

Problem Description

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.

 

Input

Standard input consists of a number of lowercase words, one per line, in alphabetical order. There will be no more than 50,000 words.
Only one case.

 

Output

Your output should contain all the hat’s words, one per line, in alphabetical order.

 

Sample Input

aahathathatwordhzieeword

 

Sample Output

ahathatword

 

Author

戴帽子的

 

大体题意：

给你许多单词，让你找出所有的单词 满足，某个单词可拆成字典里存的单词！

思路：

方法有很多，才5W个单词，UVA 10391 和这个题是一样的，可对比看，那个12W个单词！

方法1：

建立set，把所有单词存入set里，进行二分查找！

方法2，直接建立字典树，两个方法输入基本一样！

就是查找了！

方法1： set

#include<iostream>#include<string>#include<set>#include<vector>using namespace std;int main(){    set<string>s;    string p,q;    vector<string>v;    while(cin >> p){        s.insert(p);        v.push_back(p);    }    for (int i = 0; i < (int)v.size(); ++i){            int len = (int)v[i].length();        for (int k = 1; k < len; ++k){            p = v[i].substr(0,k);            q = v[i].substr(k);            set<string>::iterator it1 = s.find(p);            set<string>::iterator it2 = s.find(q);            if (it1 != s.end() && it2 != s.end()){cout << v[i] << endl;break;}        }    }    return 0;}

方法2：

字典树方法：

#include<iostream>#include<cstdio>#include<cstdlib>#include<string>#include<cstring>#include<vector>using namespace std;const int maxn = 26;struct Trie{    int id,v;    Trie *next[maxn];};Trie *root;void createTrie(const char *str){    Trie *p = root, *q;    int len = strlen(str);    for (int i = 0; i < len; ++i){        int id = str[i] - 'a';        if (p->next[id] == NULL){            q = new Trie;            for (int j = 0; j < maxn; ++j)q->next[j] = NULL;            p->next[id]= q;            p=q;        }else p=p->next[id];    }    p->v=-1;}int findTrie(const char *str){    Trie *p = root;    int len = strlen(str);    for (int i = 0; i < len; ++i){        int id = str[i] - 'a';        p=p->next[id];        if (p == NULL)return 0;    }   if (p->v == -1)return 1;    return 0;}int main(){    root = new Trie;    for (int i = 0; i < maxn; ++i)root->next[i] = NULL;    vector<string>v;    string s;    while(cin >> s){        v.push_back(s);        const char *ss = s.data();        createTrie(ss);    }    for (int i = 0; i < v.size(); ++i){        s = v[i];        int len = (int)s.length();        string a,b;        for (int k = 1; k < len; ++k){            a = s.substr(0,k);            b = s.substr(k);            const char *pp1=a.data();            const char *pp2=b.data();            if (findTrie (pp1) && findTrie(pp2)){cout << s << endl;break;}        }    }    return 0;}