02-线性结构3 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

理解要点：1.要pop n，前提是要push 1 2 3 … n-1。 num为既定序列1 2 3 … 例如，input为 4 3 2 1的4时，首先要push 1 2 3 才能push 4 pop 4 出栈。

2.stack.size() > M(the maximum capacity of the stack) 超出栈容量不可能。

思路：当栈顶元素不是input 则push（num++）直到input，随后pop input；或者stack.size() > M(the maximum capacity of the stack) 检测出不可能。

 

#include<iostream>  
#include<stack>
using namespace std;
 
int main()
{
    int M;         //maximum capacity of the stack  
    int N;         //the length of push sequence  
    int K;         //the number of pop sequence to be checked  
    cin >> M >> N >> K;
    bool flag = true; 
    int input, num;     //num= 1.2.3.4.5... input为依次输入的检测序列值 
    stack<int> sta; 
    
    for(int i = 0; i < K; i++) {
        num = 1;
        flag = true;
        for(int j = 0; j < N; j++) {
            cin >> input;  
            while( sta.size() <= M && num ) {  
                if(sta.empty() || sta.top() != input) {  
                    sta.push(num ++);  
                }else if(sta.top() == input) {  
                    sta.pop();  
                    break;  
                }             
            }  
            if(sta.size() > M )  //超出栈容量 
                flag = false; 
        } 
        if(flag)  
            cout<<"YES"<<endl;  
        else  
            cout<<"NO"<<endl;  
            
        while(!sta.empty())  //清空栈 
            sta.pop();  
    } 
    return 0; 
}

参考：http://www.2cto.com/kf/201311/254409.html　　开始误入了比大小的误区，这个比较细致易理解。记一笔。