1051. Pop Sequence (25)【栈】——PAT (Advanced Level) Practise
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题目信息
1051. Pop Sequence (25)
时间限制100 ms
内存限制65536 kB
代码长度限制16000 B
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
解题思路
用stack模拟
AC代码
#include <cstdio>#include <stack>using namespace std;int main(){ int m, n, k, t; scanf("%d%d%d", &m, &n, &k); while (k--){ stack<int> st; int used = 0; bool flag = true; for (int i = 0; i < n; ++i){ scanf("%d", &t); if (flag){ while (st.empty() || st.top() != t){ st.push(++used); if (st.size() > m){ flag = false; break; } } if (!st.empty() && st.top() == t){ st.pop(); } } } printf("%s\n", flag ? "YES":"NO"); } return 0;}
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