A1051. 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

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<stack>
using namespace std;
int main(){
    int N, M, K, temp;
    stack<int> stk;
    scanf("%d%d%d", &M, &N, &K);
    for(int i = ; i < K; i++){
        int index = , find = ;
        for(int j = ; j < N; j++){
            scanf("%d", &temp);
            if(stk.empty()){
                stk.push(index++);
            }
            if(stk.top() < temp){
                while(stk.top() < temp && stk.size() < M){
                    stk.push(index++);
                }
                if(stk.top() < temp)
                    find = ;
            }
            if(stk.top() > temp){
                find = ;
                continue;
            }
            if(stk.top() == temp){
                stk.pop();
                continue;
            }
        }
        if(find == )
            printf("NO\n");
        else printf("YES\n");
        while(!stk.empty()){
            stk.pop();
        }
    }
    cin >> N;
    return ;
}

总结：

1、栈的模拟题。注意栈的大小有限制。

2、注意每次使用完毕后清空stack。