ava tun eth som class ... left 文件 als

原題描述：

D - Rails

Description

There is a famous railway station in PopPush City. Country there is incredibly hilly. The station was built in last century. Unfortunately, funds were extremely limited that time. It was possible to establish only a surface track. Moreover, it turned out that the station could be only a dead-end one (see picture) and due to lack of available space it could have only one track.

The local tradition is that every train arriving from the direction A continues in the direction B with coaches reorganized in some way. Assume that the train arriving from the direction A has N <= 1000 coaches numbered in increasing order 1, 2, ..., N. The chief for train reorganizations must know whether it is possible to marshal coaches continuing in the direction B so that their order will be a1, a2, ..., aN. Help him and write a program that decides whether it is possible to get the required order of coaches. You can assume that single coaches can be disconnected from the train before they enter the station and that they can move themselves until they are on the track in the direction B. You can also suppose that at any time there can be located as many coaches as necessary in the station. But once a coach has entered the station it cannot return to the track in the direction A and also once it has left the station in the direction B it cannot return back to the station.

Input

The input consists of blocks of lines. Each block except the last describes one train and possibly more requirements for its reorganization. In the first line of the block there is the integer N described above. In each of the next lines of the block there is a permutation of 1, 2, ..., N. The last line of the block contains just 0.

The last block consists of just one line containing 0.

Output

The output contains the lines corresponding to the lines with permutations in the input. A line of the output contains Yes if it is possible to marshal the coaches in the order required on the corresponding line of the input. Otherwise it contains No. In addition, there is one empty line after the lines corresponding to one block of the input. There is no line in the output corresponding to the last ``null‘‘ block of the input.

Sample Input

5
1 2 3 4 5
5 4 1 2 3
0
6
6 5 4 3 2 1
0
0

Sample Output

Yes
No

Yes

這個題目用到了棧，為此我還看了下書。不過還是蠻容易的。我就在草稿紙上畫了三個方框來表示，一個表示原本的排車順序就是遞增的那種。再一個就是表示出戰的順序。還有一個表示棧。

我們從出戰第一個和進站第一個相比較，如果相同就直接過去，如果不同，我們判斷一下棧裏面又沒有，有的話就直接上去繼續判斷。看下就把進站的車放進去，這樣直到進站沒有車，如果出站的車還沒按順序出去就說明這種出戰方式不存在。直接判斷Yes No。

要註意的是案例和出站順序都是用0來表示的。棧的頭文件#include <stack>棧的定義 stack <棧的類型> 棧的名稱 棧是先進先出，並且只能取頂端數。我們這裏用到了

top()函數：取棧的頂端元素，但不刪除。

pop()函數：刪除頂端元素。

empty()：如果棧的值為空就返回 1 否則 0 。

push()函數：在頂端加入新元素。

AC代碼：

 1 #include <iostream>
 2 #include <stack>
 3 using namespace std;
 4 const int N=1010;
 5 int main()
 6 {
 7     int n;
 8      while(cin>>n&&n)
 9      {
10          int a[n+2],b[n+1],g;
11          while(cin>>b[1])
12          {
13              g=1;
14              if(b[1]==0){cout<<endl;break;}
15              a[1]=1;a[n+1]=0;
16              for(int i=2;i<=n;i++)
17             {
18                 cin>>b[i];
19                 a[i]=i;
20             }
21             stack <int> c;
22             int i=1,j=1;
23             while(j<=n)
24             {
25                 if(a[i]==b[j]) {i++;j++;}
26                 else if(!c.empty()&&b[j]==c.top()) {j++;c.pop();}
27                     else if(i<=n) {c.push(a[i]);i++;}
28                     else {g=0;break;}
29             }
30             if(g==1) cout<<"Yes"<<endl;
31             else cout<<"No"<<endl;
32          }
33      }
34     return 0;
35 }

Week 1 # D Rails