Rails

Problem 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 coaches numbered in increasing order . 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 . 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

N described above. In each of the next lines of the block there is a permutation of The last line of the block contains just 0.

The last block consists of just one line containing 0.

Output

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 Output

Yes
No

Yes

Hint

一种不用栈的写法，是查找，用两个数组一一比对。

很容易理解的数组法：

// Copyright (c) 2017 , Shawbert. Shaw
// All rights reserved.
//
// Filename: Rails
//
// Description: Stack;Array
//
// Version:  1.0
// Created:  2017/12/20  11点37分
// Compiler:  g++
//
// Author:  孙毓霄 Shawbert
#include <stdio.h>
#include <stdlib.h>

int main()
{
  int a[1111],b[1111],i,j,k,n;
  while(scanf("%d",&n),n)
  {
      while(scanf("%d",&b[0]),b[0])
      {
          for(j = 1; j < n; ++j) scanf("%d",&b[j]);
          for(i = 1,j = 0,k = 0;i <= n && j < n; ++i,++k)
          {
              a[k] = i;
              while(a[k] == b[j])
              {
                  k--;
                  j ++;
                  if(k == -1) break; // k == -1相当于栈中元素已弹出完毕，此时要回到for循环继续入栈。
              }
          }
          if(j == n) printf("Yes\n");
          else printf("No\n");
      }
      printf("\n");
  }
  return 0;
}

而另一个则是用到栈

// Copyright (c) 2017 , Shawbert. Shaw
// All rights reserved.
//
// Filename:  Rails
//
// Description: Stack实现
//
// Version:  1.0
// Created:  2017/12/20/11点40分
// Compiler:  g++
//
// Author:  曹舒淇
#include <stdio.h>
#include <stdlib.h>

typedef struct stack
{
    int arr[1003];
    int top;
} Stack;

int push(Stack* stk,int val)
{
    stk -> top ++;
    stk -> arr[stk -> top] = val;
    return 1;
}
int pop(Stack* stk)
{
    stk -> top --;
    return 1;
}

int main()
{
    int n;
    while(scanf("%d",&n) != EOF)
    {
        while(1)
        {
            Stack stk;
            stk.top = -1;     //起初无元素
            int i,j,k = 0;
            int b[1003];
            scanf("%d",&b[0]);//这行可以写while循环  while(scanf("%d",&arr[0]),arr[0])
            if(b[0] == 0) {printf("\n");break;}
            for(i = 1; i < n; i ++)  scanf("%d",&b[i]);

           /* b[0] - b[4] = 2 4 3 5 1
                          5 4 3 2 1
                          3 2 1 5 4
                    i=    1 2 3 4 5   */
                    
            for(i = 1;i <= n;i++)
            {
                push(&stk,i);                      //与a[k] = i;同理
                while(stk.top >= 0)
                {
                    if(stk.arr[stk.top] == b[k])   //比对有无相同元素，若有，标记加1，弹出该元素。
                    {
                        pop(&stk);                 //栈与数组不同，比较的是top，所以要弹出。  
                        k ++;		 	   //相当于上面的j++；
                    }
                    else break;
                }
            }
            if(k == n) printf("Yes\n");
            else printf("No\n");

        }

    }

    return 0;
}