Trie树模板例题

Posted 2020-11-26 tyty-somnuspoppy

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 = strlen(x);
    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;
}