• 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 = 0; i < K; i++){
            int index = 1, find = 1;
            for(int j = 0; 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 = 0;
                }
                if(stk.top() > temp){
                    find = 0;
                    continue;
                }
                if(stk.top() == temp){
                    stk.pop();
                    continue;
                }        
            }
            if(find == 0)
                printf("NO
    ");
            else printf("YES
    ");
            while(!stk.empty()){
                stk.pop();
            }
        }
        cin >> N;
        return 0;
    }
    View Code

    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。

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  • 原文地址:https://www.cnblogs.com/zhuqiwei-blog/p/8530271.html
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