字典树

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 293 Accepted Submission(s): 136 

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

#include <iostream>#include <sstream>#include <string.h>#include <cstdio>#include <string>#include <cctype>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <set>#include <cmath>#define PI acos(-1)#define ll long long#define INF 0x7fffffff#define _N 50005using namespace std;char word[_N][30];struct Node {    bool is;    Node *next[30];    Node() {        is = false;        memset(next, 0, sizeof(next));    }};/// create treevoid insert(Node *root, char *s){    Node *p = root;    int i = 0;    while (s[i]) {        int j = s[i++] - 'a';        if (p->next[j] == NULL) {            p->next[j] = new Node();        }        p = p->next[j];    }    p->is = true;}int search(Node *root, char *s) {    // printf("%s\n", s);    int i = 0;    int top = 0;    int stack[1005];    Node *p = root;    while (s[i]) {        int j = s[i++] - 'a';        if (p->next[j] == NULL) return 0;        p = p->next[j];        /// save        if (p->is && s[i])            stack[top++] = i;    }    while (top) {        int flag = 1;        i = stack[--top];        p = root;        while (s[i]) {            int j = s[i++] - 'a';            if (p->next[j] == NULL) {                flag = 0;                break;            }            p = p->next[j];        }        if (flag && p->is) return 1;    }    return 0;}int main() {    // freopen("in.txt", "r", stdin);    int i = 0;    Node *root = new Node();    while (~scanf("%s", word[i])) {        insert(root, word[i]);        i++;    }    for (int j = 0; j < i; j++) {        // printf("%s", word[j]);        if (search(root, word[j])) {            printf("%s\n", word[j]);        }    }    return 0;}