1 /*
 2  Problem；
 3   Tower of Hanoi
 4
 5  Question Description:
 6   there are 3 poles and n plates,all of them on the leftmost pole at first,in this question.
 7   all of the plates have different sizes and a larger one is piled up on a smaller one.
 8   What you are expected to solve is how to move these plates from the leftmost pole to the rightmost
 9   one through a series of steps. In each step you can just move the top plate from one pole to the
10   top of another pole and each smaller plate cann't be pile up on a bigger one.
11
12  Solution:
13   step1: move all of the n-1 plates from the leftmost pole to the middle pole;
14   step2: move the n-th plate from the leftmost pole to the rightmost pole;
15   step3: move the n-1 plates from the middle pole to the rightmost pole;
16
17   obviously, step3 is a recursion of this question with a smaller scale.
18 */
19
20 /*
21 source code
22 */
23 #include<iostream>
24 #include<vector>
25 using namespace std;
26
27 struct Pole{
28     vector<int> plates;//used for storing the plates
29 }Pleft,Pmiddle,Pright;
30
31 void move(Pole &sou,Pole &des){//move the top plate from sou to des
32     if(des.plates.size()==0){
33         des.plates.push_back(sou.plates[0]);
34     }
35     else{
36         des.plates.insert(des.plates.begin(),sou.plates[0]);
37     }
38     sou.plates.erase(sou.plates.begin());
39 }
40 void TOH(int n,Pole &left,Pole &middle,Pole &right){//recursion function
41     if(n==0) return;
42     TOH(n-1,left,right,middle);
43     move(left,right);
44     TOH(n-1,middle,left,right);
45 }
46
47 void print(Pole p){// print the numbers of plates of the Pole p
48     cout<<"numbers of Plates of the Pole:";
49     int i=0,n=p.plates.size();
50     for(;i<n;i++){
51         cout<<p.plates[i]<<" ";
52     }
53     cout<<endl;
54 }
55
56 int main(){
57     int n;// the quantity of the plate
58     cin>>n;
59     for(int i=0;i<n;i++){
60         Pleft.plates.push_back(i+1);// initiation
61     }
62     TOH(n,Pleft,Pmiddle,Pright);
63
64     print(Pleft);
65     print(Pmiddle);
66     print(Pright);
67 }