列车调度(Train)
Description
Figure 1 shows the structure of a station for train dispatching.
Figure 1
In this station, A is the entrance for each train and B is the exit. S is the transfer end. All single tracks are one-way, which means that the train can enter the station from A to S, and pull out from S to B. Note that the overtaking is not allowed. Because the compartments can reside in S, the order that they pull out at B may differ from that they enter at A. However, because of the limited capacity of S, no more that m compartments can reside at S simultaneously.
Assume that a train consist of n compartments labeled {1, 2, …, n}. A dispatcher wants to know whether these compartments can pull out at B in the order of {a1, a2, …, an} (a sequence). If can, in what order he should operate it?
Input
Two lines:
1st line: two integers n and m;
2nd line: n integers separated by spaces, which is a permutation of {1, 2, …, n}. This is a compartment sequence that is to be judged regarding the feasibility.
Output
If the sequence is feasible, output the sequence. “Push” means one compartment goes from A to S, while “pop” means one compartment goes from S to B. Each operation takes up one line.
If the sequence is infeasible, output a “no”.
Example 1
Input
5 2
1 2 3 5 4
Output
push
pop
push
pop
push
pop
push
push
pop
pop
Example 2
Input
5 5
3 1 2 4 5
Output
No
Restrictions
1 <= n <= 1,600,000
0 <= m <= 1,600,000
Time: 2 sec
Memory: 256 MB
描述
某列车调度站的铁道联接结构如Figure 1所示。
其中，A为入口，B为出口，S为中转盲端。所有铁道均为单轨单向式：列车行驶的方向只能是从A到S，再从S到B；另外，不允许超车。因为车厢可在S中驻留，所以它们从B端驶出的次序，可能与从A端驶入的次序不同。不过S的容量有限，同时驻留的车厢不得超过m节。
设某列车由编号依次为{1, 2, ..., n}的n节车厢组成。调度员希望知道，按照以上交通规则，这些车厢能否以{a1, a2, ..., an}的次序，重新排列后从B端驶出。如果可行，应该以怎样
的次序操作?
输入
共两行。
第一行为两个整数n，m。
第二行为以空格分隔的n个整数，保证为{1, 2, ..., n}的一个排列，表示待判断可行性的驶出序列{a1，a2，...，an}。
输出
若驶出序列可行，则输出操作序列，其中push表示车厢从A进入S，pop表示车厢从S进入B，每个操作占一行。
若不可行，则输出No。
样例
见英文题面
限制
1 ≤ n ≤ 1,600,000
0 ≤ m ≤ 1,600,000
时间：2 sec
空间：256 MB
C++:
/*
@Date : 2018-02-05 18:52:19
@Author : 酸饺子 ([email protected])
@Link : https://github.com/SourDumplings
@Version : $Id$
*/
/*
https://dsa.cs.tsinghua.edu.cn/oj/problem.shtml?id=1145
*/
#include <iostream>
#include <cstdio>
using namespace std;
const int SZ = 1<<20; //快速io
struct fastio{
char inbuf[SZ];
char outbuf[SZ];
fastio(){
setvbuf(stdin,inbuf,_IOFBF,SZ);
setvbuf(stdout,outbuf,_IOFBF,SZ);
}
}io;
const int maxn = 1600001;
// const int maxn = 1600;
int S[maxn];
int top = -1;
int B[maxn];
char order[2*maxn];
int o = 0;
int main(int argc, char const *argv[])
{
int n, m;
scanf("%d %d", &n, &m);
int topA = 1;
bool judge = true;
for (int i = 0; i != n; ++i)
scanf("%d", &B[i]);
for (int i = 0; i != n;)
{
if (top != -1 && S[top] == B[i])
{
order[o++] = 'o';
--top;
++i;
}
else
{
while (topA <= n && top < m - 1 && (top == -1 || S[top] != B[i]))
{
order[o++] = 'u';
S[++top] = topA++;
}
if ((topA > n || top >= m - 1) && S[top] != B[i])
judge = false;
}
if (!judge)
break;
}
if (judge)
{
for (int i = 0; i != o; ++i)
{
switch (order[i])
{
case 'o' : printf("pop\n"); break;
case 'u' : printf("push\n"); break;
}
}
}
else
printf("No\n");
return 0;
}