A1051

1051 Pop Sequence (25分)

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

按1~N的顺序push,可任意pop,Stack可存的元素的大小为m，判断能否按照给出的序列顺序输出。

#include<bits/stdc++.h>
using namespace std;
int main(){
    int M,N,K;
    cin>>M>>N>>K;
    for(int i=0;i<K;i++){
        stack<int>s;
        vector<int>v(N+1);
        for(int j=1;j<=N;j++)
            cin>>v[j];
        int is=0,index=1;       
        for(int k=1;k<=N;k++){
            s.push(k);
            if(s.size()>M)break;
            while(!s.empty()&&v[index]==s.top()){
                index++;
                s.pop();
            }
        }
        if(index==N+1)is=1;
        printf("%s\n",is?"YES":"NO");
    }
    return 0;
}