Trie (Prefix Tree)

What is a dictionary tree?

Trie I think more should be called a prefix tree, because to use it to seek a string prefix very convenient.

Through the above this figure, we find that:

1, with the edges represent the letter trie

2, a common prefix with the word nodes with the same prefix (It is for this reason we have to find a prefix string is very convenient)

3, the root of the whole tree is empty.

4, the end of each word is represented by a special character, figure by black $^{'} ，那么从根节点到任意一个 ‘$

Basic operation:

First, insert

Suppose we want to insert the string "in". We start at the root, i.e. node 0, P = 0 we expressed. We look at P Is there a logo with edge i connected to child nodes. Without this side, so we create a new node, i.e. node No. 1, No. 1 and the node P is set to 0 sub-node number of the node, and the edge identified as i. Finally, we move to the 1st node, that is, so P = 1.

So we put the "in" characters into the i in the Trie. Then we insert characters n, P is first find a node with a number that is not marked as the n side. Or not, then a further new node 2, is set as P 1 sub-node number of the node, and identify the side is n. And then finally moved to P = 2. So we put n also inserted. Since n is "in" the last character, so we also need to P = 2 this node is marked as the end point.

Now we insert the string "inn". The process is the same, from P = 0 i started looking logo for the side, this time to find the No. 1 node. So we do not need to create a new node, and move directly to the node No. 1, that is, so P = 1. Then insert the character n, is the presence of node No. 2, No. 2 to the mobile node, P = 2. Finally, insert the character is not identified as P n time n side, so the new node as the No. 3 No. 2 node's children, identified as the n side, while the serial number of the endpoint node 3:

Repeat, we can get the best view of the start up!

Second, look for

既然插入已经知道了，那么我们现在就来看看查找吧

查找其实比较简单。我们只要从根节点开始，沿着标识着S[1] -> S[2] -> S[3] … -> S[S.len]的边移动，如果最后成功到达一个终结点，就说明S在Trie树中；如果最后无路可走，或者到达一个不是终结点的节点，就说明S不在Trie树中。

如果是查找”te”，就会从0开始经过5最后到达6。但是6不是终结点，所以te没在Trie树中。如果查找的是”too”，就会从0开始经过5和9，然后发现之后无路可走：9号节点没有标记为o的边连出去。所以too也不在Trie中。

代码实现：

数组方式实现
要写代码实现一个Trie首先就要确定如何存储一个Trie结构。这里用一个二维数组来存储：

int trie[MAX_NODE][CHARSET];
int k;
其中MAX_NODE是trie中最大能存储的节点数目，CHARSET是字符集的大小，k是当前trie中包含有多少个节点。Trie[i][j]的值是0表示trie树中i号节点，并没有一条连出去的边，满足边上的字符标识是字符集中第j个字符（从0开始）；trie[i][j]的值是正整数x表示trie树中i号节点，有一条连出去的边，满足边上的字符标识是字符集中第j个字符，并且这条边的终点是x号节点。

具体代码：

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <string>
 4 #include <iostream>
 5 #include <stdlib.h>
 6 #include <algorithm>
 7 
 8 using namespace std;
 9 const int MAX_NODE = 300001;
10 const int CHARSET = 26;
11 
12 int trie[MAX_NODE][CHARSET]={0};
13 int color[MAX_NODE]={0};
14 int k = 1;
15 
16 void insert(char *w)
17 {
18     int len = strlen(w);
19     int p = 0;
20     for (int i=0;i<len;i++)
21     {
22         int c = w[i]-'a';
23         if (!trie[p][c])
24         {
25             trie[p][c] = k;
26             k++;
27         }
28         p = trie[p][c];
29     }
30     color[p] = 1;
31 }
32 
33 
34 int search(char *s)
35 {
36     int len = strlen(s);
37     int p = 0;
38     for (int i=0;i<len;i++)
39     {
40         int c = s[i]-'a';
41         if (!trie[p][c])
42             return 0;
43         p = trie[p][c];
44     }
45     return color[p] == 1; //是不是其中的一个
46 }
47 
48 
49 int main()
50 {
51 #ifndef ONLINE_JUDGE
52     freopen("../in.txt","r",stdin);
53 #endif
54     int t,q;
55     char s[20];
56     scanf("%d%d", &t,&q);
57     while(t--){
58         scanf("%s", s);
59         insert(s);
60     }
61     while(q--){
62         scanf("%s", s);
63         if(search(s)) printf("YES\n");
64         else printf("NO\n");
65     }
66     return 0;
67 }

参考博客：https://blog.csdn.net/weixin_39778570/article/details/81990417

　　　　 https://www.cnblogs.com/TheRoadToTheGold/p/6290732.html

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