PAT 1051. Pop Sequence

Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, ..., N and pop randomly. You are supposed to tell if a given sequence of numbers is a possible pop sequence of the stack. For example, if M is 5 and N is 7, we can obtain 1, 2, 3, 4, 5, 6, 7 from the stack, but not 3, 2, 1, 7, 5, 6, 4.

Input Specification:

Each input file contains one test case. For each case, the first line contains 3 numbers (all no more than 1000): M (the maximum capacity of the stack), N (the length of push sequence), and K (the number of pop sequences to be checked). Then K lines follow, each contains a pop sequence of N numbers. All the numbers in a line are separated by a space.

Output Specification:

For each pop sequence, print in one line "YES" if it is indeed a possible pop sequence of the stack, or "NO" if not.

Sample Input:

5 7 5
1 2 3 4 5 6 7
3 2 1 7 5 6 4
7 6 5 4 3 2 1
5 6 4 3 7 2 1
1 7 6 5 4 3 2

Sample Output:

YES
NO
NO
YES
NO

分析

对每组给的数据进行遍历，遇到一个数num，首先说明1到num都已经进堆了，要不然是不可能弹出n的，所以把1~num未进堆的所有数进堆，在进堆的过程中如果堆已满了，却依然要进堆，说明这组数据错了；用t记录进堆中最大的数，这样下次就不用从1~num中找未进堆的再进堆了，直接从t+1~到num进堆,然后pop一个出来，如果pop出来的和num不等，就错了，否则继续遍历，执行前面的动作。

#include<iostream>
#include<math.h>
#include<stack>
using namespace std;
int main(){
	int cmax,n,k,num=0;
	cin>>cmax>>n>>k;
	for(int i=0;i<k;i++){
		int flag=0,t=0;
		stack<int> s;
		for(int j=0;j<n;j++){
			cin>>num;     
		    for(int i=t+1;i<=num;i++){
                if(s.size()==cmax) flag=1;
		        s.push(i);
		    }
		    t=max(num,t);
		    if(s.top()!=num) flag=1;
		    s.pop();
		}
		if(flag==1) cout<<"NO"<<endl;
		else cout<<"YES"<<endl; 
	}
	return 0;
}