zoukankan      html  css  js  c++  java
  • HDU Problem 1247 Hat's Words 【字典树】

    Hat’s Words

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 13908    Accepted Submission(s): 4987

    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
    a ahat hat hatword hziee word
     
    Sample Output
    ahat hatword
     
    Author
    戴帽子的
     
    Recommend
    Ignatius.L   |   We have carefully selected several similar problems for you:  1075 1298 1800 2846 1305 

    题意:给你n个单词,处理到EOF。问这些单词中有几个单词是由两个拼接成的

    思路:

    用字典树,枚举每个单词,找到这个单词每个节点上的单词结尾标志,记录,然后检查剩下的单词出没出现在字典树上。

    #include <map>
    #include <set>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <queue>
    #include <iostream>
    #include <stack>
    #include <cmath>
    #include <string>
    #include <vector>
    #include <cstdlib>
    //#include <bits/stdc++.h>
    //#define LOACL
    #define space " "
    using namespace std;
    typedef long long LL;
    typedef __int64 Int;
    typedef pair<int, int> paii;
    const int INF = 0x3f3f3f3f;
    const double ESP = 1e-5;
    const double PI = acos(-1.0);
    const int MOD = 1e9 + 7;
    const int MAXN = 100 + 10;
    char str[50000 + 10][30];
    // 字典树
    struct trie {
        int cnt; //记录该节点下的单词数目
        trie* next[30]; //以每个单词创建子树
        bool endd; //该节点是否为单词的最后一位
    } *root;
    //创建根节点
    void build_trie() {
         root = new trie;
         for(int i = 0;i < 27; i++)
              root->next[i] = NULL;
    }
    void insert_trie(char s[]) {
        int len = strlen(s);
        trie *p = root, *q;
        for (int i = 0; i < len; i++) {
            int idx = s[i] - 'a';
            //如果该字母没有出现
            if (p->next[idx] == NULL) {
                q = new trie;
                q->cnt = 0;
                q->endd = false;
                for (int j = 0; j < 30; j++) {
                    q->next[j] = NULL;
                }
                p->next[idx] = q;
            }
            p = p->next[idx];
            //记录此时的单词数目
            p->cnt++;
        }
        //标记结尾
        p->endd = true;
    }
    bool query_trie(char s[]) {
        trie *p = root;
        int len = strlen(s);
        int num[50000];
        int px = 0;
        for (int i = 0; i < len; i++) {
            int idx = s[i] - 'a';
            if(p->next[idx]->endd) num[px++] = i;
            p = p->next[idx];
        }
        for (int i = 0; i < px; i++) {
            p = root;
            int j = num[i] + 1;
            for (; j < len; j++) {
                int idx = s[j] - 'a';
                if(p->next[idx] == NULL) break;
                p = p->next[idx];
                if(j == len - 1 && p->endd) return true;
            }
        }
        return false;
    }
    int main() {
        build_trie(); int t = 0;
        while (scanf("%s", str[t]) != EOF) {insert_trie(str[t]); t++;}
        for (int i = 0; i < t; i++) {
            if (query_trie(str[i])) printf("%s
    ", str[i]);
        }
        return 0;
    }



  • 相关阅读:
    Python 的编码格式
    Python import其他层级的模块
    自己写ORM框架 DBUtils_DG Java(C#的写在链接里)
    C#对象深度克隆
    SpringMVC文件上传下载
    HttpRuntime.Cache .Net自带的缓存类
    Winform跨窗体操作控件(使用委托)
    Winform调用WebKitBrowser,基于chrome内核WebKit的浏览器控件
    ORM框架 EF code first 的封装 优化一
    Go Language 开发环境搭建
  • 原文地址:https://www.cnblogs.com/cniwoq/p/6770750.html
Copyright © 2011-2022 走看看