PAT (Advanced Level) Practise 1051 Pop Sequence (25)

1051. Pop Sequence (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 51 2 3 4 5 6 73 2 1 7 5 6 47 6 5 4 3 2 15 6 4 3 7 2 11 7 6 5 4 3 2

Sample Output:

YESNONOYESNO

题意：给你n个数，让你按1~n的顺序压入栈，栈的大小为m，问能不能使得弹出顺序为给出的序列

解题思路：用stack模拟即可

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int n, m, q;int a[1009];int main(){while (~scanf("%d%d%d", &m, &n, &q)){while (q--){for (int i = 1; i <= n; i++) scanf("%d", &a[i]);stack<int>s;int cnt = 1;for (int i = 1; i <= n; i++){s.push(i);if (s.size() > m) break;while (!s.empty() && s.top() == a[cnt]) s.pop(), cnt++;}if (s.empty()) printf("YES\n");else printf("NO\n");}}return 0;}