POJ-1363 Rails-堆疊入門
Rails
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 28286 | Accepted: 10992 |
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.
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.
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
Source
Central Europe 1997
這道題的意思是,車只有兩種開的方向:A入站、站出B。
自由移動的意思是指,車子可以開進站不立即出去,也可以開進去就開出去。
要模擬出入站,看看是否滿足給定的順序。
這道題的意思是,車只有兩種開的方向:A入站、站出B。
自由移動的意思是指,車子可以開進站不立即出去,也可以開進去就開出去。
要模擬出入站,看看是否滿足給定的順序。
//AC1
#include<stdio.h>
int main()
{
int a[1005],b[1005],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])
{
if(k>0)k--;
else
{
a[k]=0,k--;
}
j++;
if(k==-1)break;
}
}
if(j==n)printf("Yes\n");
else printf("No\n");
}
printf("\n");
}
}
//AC2
#include<stdio.h>
int main()
{
int input[1005],output[1005],k,j,i,n;
while(scanf("%d",&n),n)
{
while(scanf("%d",&output[0]),output[0])
{
for(j=1; j<n; j++)
scanf("%d",&output[j]);
for(k=1,j=0,i=0; k<=n;i++,k++)
{
input[i]=k;
while(input[i]==output[j])
{
i--;
j++;
if(i==-1)
break;
}
}
if(j==n)
printf("Yes\n");
else
printf("No\n");
}
printf("\n");
}
}
//超時
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int maxn=1000+10;
int main()
{
int n,p[maxn];
cin>>n;
while(n)
{
int x,max=0,i,j;
cin>>x;
while(x)
{
memset(p,0,sizeof(p));
bool valid=true;
for(i=0; i<n; ++i)
{
if(valid)
{
bool ok=true;
for(j=x+1; j<=max; ++j)
if(p[j]==1)
{
ok=false;
break;
}
if(!ok)
valid=false;
else
{
max=(max>x?max:x);
p[x]=2;
for(j=x-1; j>0&&!p[j]; --j)
p[j]=1;
}
}
if(i<n)
cin>>x;
}
cout<<(valid?"Yes":"No")<<endl;
cin>>x;
}
cout<<endl;
cin>>n;
}
}
相關文章
- JS 堆疊JS
- java堆疊Java
- 圖的深度優先遍歷[非堆疊、堆疊實現]
- python資料視覺化-matplotlib入門(5)-餅圖和堆疊圖Python視覺化
- 記憶體堆疊記憶體
- 堆疊的工作原理
- C#堆疊(Stack)C#
- C#中堆和堆疊的區別C#
- [golang]如何看懂呼叫堆疊Golang
- 華為裝置堆疊原理
- Thrift的網路堆疊
- C++堆疊詳解C++
- 泛型鏈式堆疊泛型
- 第六講 堆疊操作
- 益智補劑:Stamets堆疊
- 圖的深度優先遍歷(堆疊實現和非堆疊實現)
- junkman 遠端堆疊監控
- 什麼是網路堆疊?
- Java 堆疊記憶體分配Java記憶體
- iOS crash 日誌堆疊解析iOS
- (js佇列,堆疊) (FIFO,LIFO)JS佇列
- z-index堆疊規則Index
- 作業系統---在核心中重新載入GDT和堆疊作業系統
- JS 資料型別和堆疊JS資料型別
- CSS之定位和堆疊屬性CSS
- SQL隱碼攻擊-堆疊注入SQL
- Java堆疊的區別有哪些Java
- QT分局管理:堆疊窗體(三)QT
- 如何:強化TCP/IP堆疊安全TCP
- 【原創】命令堆疊(二十七)
- android I/DEBUG堆疊資訊Android
- 虛擬地址空間,堆疊,堆,資料段,程式碼段
- C中關於堆疊的總結
- information_schema.innodb_trx 查詢堆疊ORM
- 【matplotlib 實戰】--堆疊面積圖
- Python實現堆疊與佇列Python佇列
- 如何強化TCP/IP 堆疊安全教程TCP
- java 堆疊的使用方法說明Java