hdu 1075 What Are You Talking About 字典樹 或 stl map

題意： 給出一本字典， 和一篇文章，根據字典翻譯文章，若字典裏面沒有文章中的某個單詞，則不用翻譯

解法：兩種解法： 1. 用c++ 的stl裏面的map來求解， 直接把每個單詞用map對應即可

2. 根據生詞表建立字典樹， 直接在字典樹裏面查詢即可， 下面簡要介紹一下字典樹：

字典樹： 以英文小寫字母爲例： 一個字典樹有一個根節點：root， 根節點下面有26個節點，分別對應a~z ,此後，從根節點往下，每個節點下面均有26個節點即a~z, 這樣，一個單詞，如 love, 順着根節點往下，依次是：next[11], next[14], next[21], next[4], 則一共經歷了4層，這樣查詢時間複雜度僅爲o(len), (len爲單詞長度）。這裏給出動態分配內存的字典樹代碼，根據代碼很好寫出靜態分配內存的代碼：

CODE of c++ stl map: 1546ms

#include <map>
#include <iostream>
#include <cstdio>
#include <string>
using namespace std;
map<string, string> dic;

int main()
{
   // freopen("/Users/apple/Desktop/in.txt", "r", stdin);
    string word1, word2, word;
    cin >> word;
    while (cin >> word1 && word1 != "END")
    {
        cin >> word2;
        dic[word2] = word1;
    }
    cin >> word;
    getchar();
    while (getline(cin, word) && word != "END")
    {
        word1 = "";
        for (int i = 0; i < word.size(); i++) {
            if (word[i] >= 'a' && word[i] <= 'z') {
                word1 += word[i];
            }
            else {
                if (dic.find(word1) == dic.end()) cout << word1;
                else cout << dic[word1];
                cout << word[i];
                word1 = "";
            }
        }
        printf("\n");
    }
    
    return 0;
}

CODE of Tree Tree: 578ms

#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
using namespace std;
const int maxn = 26;
typedef string Type;
struct Tree {
    Type s;
    Tree* next[maxn];
};
typedef Tree* pTree;
pTree root;

void init()
{
    root = new Tree;
    for (int i = 0; i < maxn; i++) {
        root->next[i] = NULL;
    }
    root->s = "";
}
void insert(string a, string b)
{
    pTree p = root, q;
    for (int i = 0; i < a.size(); i++) {
        int id = a[i] - 'a';
        if (p->next[id] == NULL) {
            q = new Tree;
            q->s[0] = 0;
            for (int j = 0; j < maxn; j++) {
                q->next[j] = NULL;
            }
            p->next[id] = q;
        }
        p = p->next[id];
    }
    p->s = b;
}
string find(string a)
{
    pTree p = root;
    for (int i = 0; i < a.size(); i++) {
        int id = a[i] - 'a';
        if (p->next[id] == NULL) {
            return a;
        }
        p = p->next[id];
    }
    if (p->s[0]) {
        return p->s;
    }
    return a;
}
void clear(pTree u)
{
    for (int i = 0; i < maxn; i++) {
        if (u->next[i]) {
            clear(u->next[i]);
        }
    }
    free(u);
}

int main()
{
    init();
//    freopen("/Users/apple/Desktop/in.txt", "r", stdin);
    string word1, word2, word;
    cin >> word;
    while (cin >> word1 && word1 != "END")
    {
        cin >> word2;
        insert(word2, word1);
    }
    cin >> word;
    getchar();
    while (getline(cin, word) && word != "END")
    {
        word1 = "";
        
        for (int i = 0; i < word.size(); i++) {
            if (word[i] >= 'a' && word[i] <= 'z') {
                word1 += word[i];
            }
            else {
                cout << find(word1);
                cout << word[i];
                word1 = "";
            }
        }
        printf("\n");
    }
    clear(root);
    
    return 0;
}