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  • Trie树模板例题

    一、模板

    结构体

    struct Trie{
        int val;
        Trie *nex[26];
        Trie(){
            val = 0;
            for(int i = 0; i < 26; ++i) nex[i] = NULL;
        }
    };
    

    建树

    void build(string x, Trie *root){
        int len = x.size();
        for(int i = 0; i < len; ++i){
            int cur = x[i] - 'a';
            if(root -> nex[cur] == NULL){
                root -> nex[cur] = new Trie();
            }
            root = root -> nex[cur];
            ++root -> val;
        }
    }
    

    查询

    int query(string x, Trie *root){
        int len = x.size();
        for(int i = 0; i < len; ++i){
            int cur = x[i] - 'a';
            if(root -> nex[cur] == NULL) return 0;
            root = root -> nex[cur];
        }
        return root -> val; //val存的是包含某前缀的字符串数
    }
    

    二、简单例题

    (1)HDU 1251

    题意:统计以某个字符串为前缀的单词数量(单词本身也是自己的前缀)。单词只有小写字母组成,不会有重复的单词出现。

    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<cmath>
    #include<algorithm>
    #include<string>
    #include<iostream>
    #include<set>
    #include<map>
    #include<stack>
    #include<queue>
    #include<vector>
    #include<sstream>
    typedef long long LL;
    const int INF = 0x3f3f3f3f;
    using namespace std;
    const int MAXN = 100000 + 10;
    const double eps = 1e-8;
    int dcmp(double a, double b){
        if(fabs(a - b) < eps) return 0;
        return a < b ? -1 : 1;
    }
    struct Trie{
        int val;
        Trie *nex[26];
        Trie(){
            val = 0;
            for(int i = 0; i < 26; ++i) nex[i] = NULL;
        }
    };
    char s[20];
    void build(Trie *root){
        int len = strlen(s);
        for(int i = 0; i < len; ++i){
            int cur = s[i] - 'a';
            if(root -> nex[cur] == NULL){
                root -> nex[cur] = new Trie();
            }
            root = root -> nex[cur];
            ++root -> val;
        }
    }
    int query(Trie *root){
        int len = strlen(s);
        for(int i = 0; i < len; ++i){
            int cur = s[i] - 'a';
            if(root -> nex[cur] == NULL) return 0;
            root = root -> nex[cur];
        }
        return root -> val;
    }
    int main(){
        Trie *root = new Trie();
        while(gets(s) && s[0] != ''){
            build(root);
        }
        while(scanf("%s", s) == 1){
            printf("%d
    ", query(root));
        }
        return 0;
    }
    

    (2)KickStart A轮 D题:https://www.cnblogs.com/tyty-Somnuspoppy/p/12546794.html

    题意:有N个字符串,每个字符串只包含字母A~Z,要求把字符串按K个一组进行分组,每个字符串只能分到一个组中。分好后,每个组的score为该组所有字符串的最长公共前缀。问分组后每个组的score之和的最大值。

    (3)HDU 2072

    题意:统计一篇文章里不同单词的总数。

    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<cmath>
    #include<algorithm>
    #include<string>
    #include<iostream>
    #include<set>
    #include<map>
    #include<stack>
    #include<queue>
    #include<vector>
    #include<sstream>
    typedef long long LL;
    const int INF = 0x3f3f3f3f;
    using namespace std;
    const int MAXN = 1000000 + 10;
    const double eps = 1e-8;
    int dcmp(double a, double b){
        if(fabs(a - b) < eps) return 0;
        return a < b ? -1 : 1;
    }
    struct Trie{
        int val;
        Trie *nex[26];
        Trie(){
            val = 0;
            for(int i = 0; i < 26; ++i) nex[i] = NULL;
        }
    };
    int ans;
    void build(string x, Trie *root){
        int len = x.size();
        for(int i = 0; i < len; ++i){
            int cur = x[i] - 'a';
            if(root -> nex[cur] == NULL){
                root -> nex[cur] = new Trie();
            }
            root = root -> nex[cur];
        }
        if(root -> val == 0){
            ++ans;
            root -> val = 1;
        }
    }
    string s;
    int main(){
        while(getline(cin, s) && s != "#"){
            Trie *root = new Trie();
            stringstream ss(s);
            string x;
            ans = 0;
            while(ss >> x){
                build(x, root);
            }
            printf("%d
    ", ans);
        }
        return 0;
    }
    

     

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  • 原文地址:https://www.cnblogs.com/tyty-Somnuspoppy/p/12552004.html
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