02-线性结构4 Pop Sequence（25 分）

02-线性结构4 Pop Sequence（25 分）

标签（空格分隔）： 数据结构 C++

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02-线性结构4 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

看了半天才看明白，其实和当初做的那道经典的UVA514 Rail是一样的。不一样的地方就是现在的栈有最大容量。只需要写一个栈的模拟就行了。
此外，还需要注意的就是边界条件了。这里的边界条件，就是程序跳出循环的点。YES的条件很好找，只需要i遍历全部元素就行了。但是NO呢？这时考虑到，由于每次元素都需要进栈，之后再判断该元素是否等于目标元素，所以只需要判断栈空间满了的时候，栈顶元素是否等于目标序列对应元素就行了，如果不等，就输出NO。

#include <iostream>
#include <algorithm>
#include <stack>
#define N 1003
using namespace std;
int main()  {
    int m,n,k;
    scanf("%d %d %d",&m,&n,&k);
    while (k--) {
        int aim[N],i=1,j=1;
        stack<int> st;
        for (int ii=1;ii<=n;ii++)
            scanf("%d",&aim[ii]);
        while (1)   {
            if (st.empty()) {st.push(j);j++;}
            else if (!st.empty()&&st.top()==aim[i]) {st.pop();i++;}
            else    {st.push(j);j++;}
            if (i>n)    {printf("YES\n");break;}
            else if (st.size()>=m)  {
                if (st.top()!=aim[i])   {
                    printf("NO\n");
                    break;
                }
            }
        }
    }
    return 0;
}