本身由于初学，只是为了练习stack的使用，基本的思路还是参照刘汝佳书中的想法，尚不具备独创性。
题目如下：
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 file 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 file contains the lines corresponding to the lines with permutations in the input file. A line of the output file contains ‘Yes’ if it is possible to marshal the coaches in the order required on the corresponding line of the input file. Otherwise it contains ‘No’. In addition, there is one empty line after the lines corresponding to one block of the input file. There is no line in the output file corresponding to the last “null” block of the input file.
Sample Input
5
1 2 3 4 5
5 4 1 2 3
0
6
6 5 4 3 2 1
0
0
Sample OutputYes
No
Yes
 |
| 火车铁轨如图所示 |
由于C站有明显的后进先出（Last In First Out，LIFO）特点，因此它是一个栈。
用A和B分别表示车厢的顺序：A表示在车站A最外侧即将发动的车厢的编号，而B表示在车站B下一辆需要到达的车厢编号。而数组target则用来储存需要被判断是否可行的到达顺序。
当B中需要编号为 i 的车厢时，能够满足的情况有两种，一是A的最前方是车厢 i；第二种是栈C最顶端是这节车厢 i。如果不满足，那么车厢 i 就只有压入中转站C里等待了。
一个要注意的问题是输出格式，每一个输入都对应一行，如果是输入0，应该对应一个空行。
则代码如下：
当B中需要编号为 i 的车厢时，能够满足的情况有两种，一是A的最前方是车厢 i；第二种是栈C最顶端是这节车厢 i。如果不满足，那么车厢 i 就只有压入中转站C里等待了。
一个要注意的问题是输出格式，每一个输入都对应一行，如果是输入0，应该对应一个空行。
则代码如下：
#include <cstdio>#include <stack> using namespace std; int main(void) { int n,target[1005] = {0}; while(scanf("%d",&n) == 1 && n){ stack<int> s; while(1){ int A = 1, B = 1; for(int i = 1; i<=n; i++){ scanf("%d",target+i); if(target[1] == 0) break; } if(target[1] != 0){ bool flag = true; while(B <= n){ if(A == target[B]){ A++,B++; } else if(!s.empty() && s.top() == target[B]){ s.pop(); B++; } else if(A <= n) s.push(A++); else{ flag = false; break; } } printf("%s\n", flag? "Yes" :"No"); } else { printf("\n"); break; } } } return 0; }
