题目链接：http://codeforces.com/gym/102222/problem/E
In computer science, a 2-3-4 tree, as a search tree, is a self-balancing data structure that is commonly used to implement dictionaries. The numbers mean a tree where every node with children (internal node) has either two, three or four child nodes.
A 2-node is the same as a binary node, which has one data element, and if internal has two child nodes. A 3-node has two data elements, and if internal has three child nodes. A 4-node has three data elements, and if internal has four child nodes. All leaves are at the same depth, and all data is kept in sorted order.
Here we introduce the procedure of inserting a value into a 2-3-4 tree. If the number of values that have been inserted in is no more than 22, the 2-3-4 tree after the insertion has only one node (the root) with several data elements.
Otherwise, we start the following procedure at the root of the 2-3-4 tree.
- If the current node is a 4-node:
- Remove and save the middle value to get a 3-node
- Split the remaining 3-node up into a pair of 2-nodes (the now missing middle value is handled in the next step).
- If this is the root node (which thus has no parent):
- the middle value becomes the new root 2-node and the tree height increases by 1. Ascend into the root (which means that the current node becomes the new root node).
- Otherwise, push the middle value up into the parent node. Ascend into the parent node (which means that the current node becomes its parent node).
- If the current node is a leaf, insert the value into the node and finish.
- Otherwise,
- Find the child whose interval contains the value to be inserted.
- Descend into the child and repeat from step 1.
Now you are given a 2-3-4 tree which is empty at first. Then we insert nn values a1,a2,⋯,ana1,a2,⋯,an into the 2-3-4 tree one by one. You are asked to write a program to print the 2-3-4 tree after inserting all the given values along the pre-order traversal. For a node with one or more data elements, your program will print these data elements in one line in ascending order.
The input contains several test cases, and the first line is a positive integer TT indicating the number of test cases which is up to 5050.
For each test case, the first line contains an integer n (1≤n≤5000)n (1≤n≤5000) indicating the number of values that will be inserted into the 2-3-4 tree. The second line contains nn integers a1,a2,⋯,ana1,a2,⋯,an. We guarantee that a1,a2,⋯,ana1,a2,⋯,an is a permutation of 1,2,⋯,n1,2,⋯,n.
For each test case, output a line containing Case #x: at first, where x is the test case number starting from 11. The following several lines describe the 2-3-4 tree along the pre-order traversal, each line of which describes a node with several integers indicating the data elements of a node in the 2-3-4 tree.
3
4
1 2 3 4
4
4 3 2 1
17
6 3 5 7 1 10 2 9 4 8 11 12 13 14 15 16 17
Case #1:
2
1
3 4
Case #2:
3
1 2
4
Case #3:
5 9
2
1
3 4
7
6
8
11 13 15
10
12
14
16 17
题目大意：就是要你要找2-3-4数的插入方法，并且最后按照前序遍历的方法输出每个节点的数据
思路：第一次接触2-3-4树，看了挺久的，到现在也不是很懂这个代码，可能自己拍还是拍不出来吧，先放上来好了，有机会接触这种题再复习
看代码：
#include<iostream> #include<algorithm> #include<stack> #include<cstdio> #include<map> #include<queue> #include<cstring> using namespace std; typedef long long LL; const int maxn=1e5+5; int a[maxn]; int cnt,rt; struct Node { int f; vector<int>d,ch; void init(int fa,int x)//f是该节点的父亲？ x是当前的结点拥有的值？ { f=fa; d.clear(),ch.clear(); d.push_back(x);//当前节点拥有多少的所有的值 } }tr[maxn]; void ins(int p,int x)//(1,x) { // cout<<"rt:"<<rt<<endl; if(tr[p].d.size()==3)//代表有三个元素？？ 也就是当前节点插满了不能再插进元素了 { int f=tr[p].f;//父亲是谁 vector<int>cpy_d,cpy_ch; cpy_d.swap(tr[p].d),cpy_ch.swap(tr[p].ch);//swap ?? 这时候cop_d 和cpy_ch拥有了ty[p].d和ty[p].ch的内容 而这两个变为空？ if(p==rt)//如果p和rt相等？？代表已经是根节点了 上面没有结点 { //相当于中间结点往上提 0 1 2三个结点 中间的自然是1 变为其它两个节点的父亲 tr[rt=++cnt].init(0,cpy_d[1]);//新的结点 父亲变为0 插入d[1]? tr[++cnt].init(rt,cpy_d[0]);//新的结点 父亲变为rt 插入d[0]? 相当于成为左儿子 tr[p].init(rt,cpy_d[2]);//原来的结点父亲是rt 只剩下d[2]? 相当于成为右儿子 tr[rt].ch.push_back(cnt),tr[rt].ch.push_back(p);//把这两个结点加入到他们的父亲节点的孩子中 if(cpy_ch.size())//如果当前节点有孩子 自然孩子也需要进行操作 { //左边两个分到左边 右边两个分到右边 大概是为了平衡吧。。。 tr[cnt].ch.push_back(cpy_ch[0]),tr[cpy_ch[0]].f=cnt;//原来的左儿子自然也就成为现在左儿子的左儿子了 tr[cnt].ch.push_back(cpy_ch[1]),tr[cpy_ch[1]].f=cnt;//为什么把1也归为左儿子的儿子？ tr[p].ch.push_back(cpy_ch[2]),tr[cpy_ch[2]].f=p;//为什么把2 3归为右儿子的儿子？ tr[p].ch.push_back(cpy_ch[3]),tr[cpy_ch[3]].f=p; // cout<<cpy_ch[0]<<" "<<cpy_ch[1]<<" "<<cpy_ch[2]<<" "<<cpy_ch[3]<<endl; } p=rt;//现在的根节点已经变为rt了 } else//不是根节点 上面有节点 { tr[p].init(f,cpy_d[0]);//左边结点 tr[++cnt].init(f,cpy_d[2]);//新建右边结点 tr[f].d.push_back(cpy_d[1]);//把中间结点提上去 sort(tr[f].d.begin(),tr[f].d.end());//提上元素之后显然要排序 tr[f].ch.push_back(cnt);//多加了个孩子 for(int i=tr[f].ch.size()-1;i>1;i--)//？？为什么要交换？ 懂了 因为你要保证孩子也是有序的呀 所以要把新插入的结点换到p之后 { if(tr[f].ch[i-1]!=p) swap(tr[f].ch[i-1],tr[f].ch[i]); else break; } if(cpy_ch.size()) { tr[p].ch.push_back(cpy_ch[0]),tr[cpy_ch[0]].f=p; tr[p].ch.push_back(cpy_ch[1]),tr[cpy_ch[1]].f=p; tr[cnt].ch.push_back((cpy_ch[2])),tr[cpy_ch[2]].f=cnt; tr[cnt].ch.push_back(cpy_ch[3]),tr[cpy_ch[3]].f=cnt; } p=f; } } // cout<<"*"<<tr[p].d.size()<<endl; //保证当前节点只有两个元素??错的 会有三个 if(tr[p].ch.size()==0)//没有孩子?? 也就不需要在孩子中进行比较和插入了 直接放进去即可 { tr[p].d.push_back(x);//作为本节点的元素 sort(tr[p].d.begin(),tr[p].d.end());//你必须保证里面的有序的 所有每次插入元素都需要排序？？？ } else//有孩子 插入元素就必须和孩子比较了 { if(x<tr[p].d[0]) ins(tr[p].ch[0],x);//如果x<第一个元素 那么很显然 肯定要插入到做儿子中 （因为元素是有序的）所以递归左儿子 else if(x>tr[p].d[tr[p].d.size()-1]) ins(tr[p].ch[tr[p].ch.size()-1],x);//如果大于最后一个元素 显然是插到右孩子中的 递归右儿子 else//不比第一个小 不比最后一个大 则肯定插入到中间 { for(int i=1;i<tr[p].d.size();i++)//遍历中间元素 if(x<tr[p].d[i]) {ins(tr[p].ch[i],x);break;}//小于谁就插入到谁的孩子中 } } } void dfs(int p) { for(int i=0;i<tr[p].d.size();i++) printf("%d%c",tr[p].d[i],i==tr[p].d.size()-1?' ':' '); for(int i=0;i<tr[p].ch.size();i++) dfs(tr[p].ch[i]); } int main() { int T; cin>>T; int ca=1; while(T--) { int N; cin>>N; for(int i=1;i<=N;i++) cin>>a[i]; cnt=rt=1,tr[rt].init(0,a[1]); //第一个点的父亲为0 拥有元素a[1] for(int i=2;i<=N;i++) ins(rt,a[i]);//rt不是一直是1吗？ printf("Case #%d: ", ca++); dfs(rt); } return 0; }