zoukankan      html  css  js  c++  java
  • Rails (栈)

     

    这段好像WA
    #include<cstdio>
    #include<algorithm>
    #include<iostream>
    #include<cstring>
    #include<stack>
    using namespace std;
    stack<int> C;
    int rail[1000],n;
    bool YesorNo()
    {
        int a = rail[0], b = 0;
        for (int i = 0; i < n; i++)
        {
            if (a <= rail[i])
            {
                a = rail[i];
                for (; b < a; b++)
                {
                    C.push(b+1);
                }
                b = a ;
                C.pop();
            }
            else if (a > rail[i])
            {
                if (rail[i] == C.top())
                    C.pop();
                else return false;
            }
            if (i == n - 1 && C.size() == 0)
                return true;
        }
    }
    int main()
    {
        while (cin >> n )
        {
            if (n == 0) break;
                while (1)
                {
                    cin >> rail[0];
                    if (rail[0] == 0 && getchar() == '
    ')
                    {
                        printf("
    ");
                        break;
                    }
                    for (int i = 1; i < n; i++)
                    {
                        cin>>rail[i];
                    }
                    printf(YesorNo() ? "Yes
    " : "No
    ");
                }
        }
    }

     

    AC代码
    #include<iostream>
    #include <algorithm>
    #include<stdio.h>
    #include<stack>
    #include<string>
    using namespace std;
    int main()
    {
    	int n;
    	while (cin >> n)
    	{
    		if (n == 0)
    			break;
    		int train[200][200];
    		for (int i = 0;;i++)
    		{
    			cin >> train[i][1];
    			if (train[i][1] == 0)
    				break;
    			for (int j = 2;j <= n;j++)
    				cin >> train[i][j];
    			bool doit = true;
    			stack<int> s;
    			int a = 1, b = 1;
    			while (b <= n)
    			{
    				if (a == train[i][b])
    				{
    					a++;
    					b++;
    				}
    				else if (!s.empty() && s.top() == train[i][b])
    				{
    					s.pop();
    					b++;
    				}
    				else if (a <= n)
    				{
    					s.push(a++);
    				}
    				else
    				{
    					doit = false;
    					break;
    				}
    			}
    			string str = doit ? "Yes" : "No";
    			cout << str << endl;
    		}
    		cout << endl;
    	}
    	return 0;
    }
    

      

    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 Output

    Yes

    No

    Yes

  • 相关阅读:
    Windows server 2016 解决“无法完成域加入,原因是试图加入的域的SID与本计算机的SID相同。”
    Windows Server 2016 辅助域控制器搭建
    Windows Server 2016 主域控制器搭建
    Net Framework 4.7.2 覆盖 Net Framework 4.5 解决办法
    SQL SERVER 2012更改默认的端口号为1772
    Windows下彻底卸载删除SQL Serever2012
    在Windows Server2016中安装SQL Server2016
    SQL Server 创建索引
    C#控制台或应用程序中两个多个Main()方法的设置
    Icon cache rebuilding with Delphi(Delphi 清除Windows 图标缓存源代码)
  • 原文地址:https://www.cnblogs.com/edych/p/7212124.html
Copyright © 2011-2022 走看看