POJ 1363Rails

描述

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.

输入

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.

输出

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.

样例输入

51 2 3 4 55 4 1 2 3066 5 4 3 2 100

样例输出

YesNoYes

题目解析：

   栈操作入门的经典题目，这个题目的思路就是判断出栈的顺序是否合法。

   1.规律法：

             对于出栈序列中的每一个数字，在它后面的、比它小的所有数字，一定是按递减顺序排列的。

   2.模拟法：

        1）如果当前栈为空，且入栈序列不空，则，入栈序列的下一个元素入栈；

        2）如果当前栈顶元素不等于出栈序列的首元素，那么入栈序列一直入栈，知道入栈序列为空。

        3）如果当前栈顶元素等于出战序列的首元素，那么栈顶元素弹出，出栈序列第一个元素移走；

        4 ) 如果入栈序列为空，出栈序列第一个元素仍然不等于栈顶元素，则表示2个序列是不匹配的。

附上模拟法的代码这个思想很关键，受用了，以前真的没有想到。

#include <cstdio>  #include <stack>  using namespace std;  const int maxn=1000+5;  int a[maxn];  int main()  {      int n,i,k;      while(scanf("%d",&n)&&n)      {          stack<int>s;//设立一个栈储存按顺序进栈的序列 （一个空栈）          while(scanf("%d",&a[0])&&a[0])          {              for(i=1;i<n;i++)                  scanf("%d",&a[i]);//要进行判断出栈的序列              for(i=1,k=0;i<=n;i++)              {                  s.push(i);//进栈                  while(s.top()==a[k])//判断栈顶元素和a数组是否相等                  {                      if(!s.empty()) s.pop();//栈不为空就出栈                      k++;//判断下一个位置                      if(s.empty()) break;//直到栈空就结束循环                  }              }              if(k==n) printf("Yes\n");//完全匹配就输出yes             else printf("No\n");          }          printf("\n");      }      return 0;  }