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Pop Sequence

Word count: 502Reading time: 2 min
2019/07/04 Share

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:

1
2
3
4
5
6
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:

1
2
3
4
5
YES
NO
NO
YES
NO

思路

题目大概意思是有个容量限制为m的栈,分别把1,2,3,…,n入栈,给出一个系列出栈顺序,问这些出栈顺序是否可能

先将输入的序列存到数组v中,然后按1-n的顺序把数字压进栈中,若栈超过最大范围则Break.

如果没超过,设置cur=1,从数组的第一个数字开始,看看是否与栈顶元素相等,设置一个循环若相等则一直弹出栈且cur++,若不相等就继续按顺序压入栈中,最后根据cur是否等于n+1来判断是否Yes或者No

答案

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#include <iostream>
#include <stack>
#include <vector>
using namespace std;
int main() {
int m, n, k;
scanf("%d %d %d", &m, &n, &k);
for(int i = 0; i < k; i++) {
bool flag = false;
stack<int> s;
vector<int> v(n + 1);
for(int j = 1; j <= n; j++)
scanf("%d", &v[j]);
int cur = 1;
for(int j = 1; j <= n; j++) {
s.push(j);
if(s.size() > m) break;
while(!s.empty() && s.top() == v[cur]) {
s.pop();
current++;
}
}
if(cur == n + 1) flag = true;
if(flag) printf("YES\n");
else printf("NO\n");
}
return 0;
}
CATALOG
  1. 1. Pop Sequence
    1. 1.1. 思路
    2. 1.2. 答案