PAT 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

注意最后判断时候还要判断栈是否为空，因为前面符合，后面不符合，栈里还有元素（没有溢出的情况），应该输出false。

#include<cstdio>
#include<stack>
using namespace std;

const int maxn = 1010;
int a[maxn];
stack<int> st;

int main() {
    int m, n, k;
    scanf("%d %d %d", &m, &n, &k);
    while (k--) {
        for (int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
        }
        while (!st.empty()) {
            st.pop();
        }
        bool flag = true; 
        int j = 1;
        for (int i = 1; i <= n; i++) {
            st.push(i);
            if (st.size() > m) {
                flag = false;
                break;
            }
            while (!st.empty() && st.top() == a[j]) {
                st.pop();
                j++;
            }
        }
        if (flag == true && st.empty()) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

猜你喜欢