【Hihocoder】1014 : Trie树

问题：http://hihocoder.com/problemset/problem/1014

给定一个字符串字典dict，输入字符串str, 要求从dict中找出所有以str为前缀的字符串个数。

构建Trie树：

1) 一个节点有多个子节点。用vector<Node*> nexts 存储。

2) 两个字符串的相同前缀部分共用同一条路径。

3) 每个节点包含计数变量 cnt， 表示有多少个字符串共用该节点

复杂度分析：

假定dict中的字符串个数 n, 字符串的长度 l

1）建树：O(n^2*l)

2) 查找：O(n*l)

源码： 

#include <iostream>
#include <vector>
#include <string>
using namespace std;

//node structure in Trie tree
struct Node
{
  Node(char _chr = -1){chr = _chr; cnt = 0;}
  char chr;
  vector<Node*> nexts;
  int cnt;//how many strings pass current node
};

int main()
{
    Node* root = new Node;//root of dictornary
    Node* current;
    vector<Node*> v_current;
    int cnt, vsize, i, j, strLen;
    string str;
    
    //create Trie tree
    cin>>cnt;
    while(cnt-- > 0)
    {
        cin>>str;
        strLen = str.length();
        current = root;
        for(i = 0; i < strLen; i++)
        {
            v_current = current->nexts;
            vsize = v_current.size();
            //try to find str[i] in current tree
            for(j = 0; j < vsize; j++)
            {
                if(v_current[j]->chr == str[i])
                    break;
            }
            if(j == vsize)//not found, create a new node
            {
                Node* tmp = new Node(str[i]);
                current->nexts.push_back(tmp);
            }
            current = current->nexts[j];
            current->cnt++;//increase count of current node
        }
    }
        
    cin>>cnt;
    while(cnt-- > 0)
    {
        cin>>str;
        strLen = str.length();
        current = root;
        for(i = 0; i < strLen; i++)
        {
            v_current = current->nexts;
            vsize = v_current.size();
            for(j = 0; j < vsize; j++)
                if(v_current[j]->chr == str[i])
                    break;
            if(j == vsize)//not found
            {
                cout<<'0'<<endl;
                break;
            }
            current = v_current[j];
        }
        if(i == strLen)
            cout<<current->cnt<<endl;
    }
    return 0;
}