02-线性结构4 Pop Sequence

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给定一个最大容量为M的栈，按1到N的顺序依次压入N个数字，然后随机弹出。任务是判断一系列给定的数字序列是否可能是该栈的弹出序列。输入包含M、N和K，K个待检查的弹出序列。对于每个序列，如果可能则输出'YES'，否则输出'NO'。

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.

#include <iostream>
#include <stack>
#include <vector>

using namespace std;

//pop sequences对的返回true, 否则返回false
bool dealPopSequences(const vector<int> &num, int m);    //m为栈的长度

int main()
{
    //接收M, N, K
    int M, N, K;
    cin >> M >> N >> K;
    vector<int> num;
    int total;
    while(K--){   //K条要check的pop sequences
        //接收一条pop sequences
        total = N;
        num.clear();    //清空clear
        while(total--){
            int tmp;
            cin >> tmp;
            num.push_back(tmp);
        } 
        //处理pop sequences
        bool flg = dealPopSequences(num, M);
        //处理输出
        if(flg) cout << "YES";
        else cout << "NO";
        if(K != 0) cout << endl;
    }
    
    
    
    return 0;
}

//pop sequences对的返回true, 否则返回false
bool dealPopSequences(const vector<int> &num, int m)    //m为栈的长度
{
    int n = num.size();   //n应该等于N
    stack<int> stk;
    int j = 0;
    for(int i = 1; i <= n; i++){
        if(stk.size() == static_cast<unsigned int>(m)) return false;
        else stk.push(i);
            //cout << stk.top() << " " << num[j] << endl;
        while(!stk.empty() && stk.top() == num[j]){
            //cout << stk.top() << " " << num[j] << endl;
            stk.pop();
            j++;
        }
    }
    if(j == n) return true;
    return false;
}