PAT 1051 Pop Sequence[栈][难]
1051 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
题目大意:给出一个栈最大容量,并且给出最大的N,要求是1....N是这样顺序入栈,输入K个检查的序列,需要判读是不是可能的弹出序列。
//我看见其实以为自己会,其实是不会的,以前见过这种题的,但是应该当时也没理解吧。
代码转自:https://www.liuchuo.net/archives/2232
#include <iostream>
#include <stack>
#include <vector>
#include<cstdio>
using namespace std;
int main() {
int m, n, k;
scanf("%d %d %d", &m, &n, &k);
for(int i = ; i < k; i++) {
bool flag = false;
stack<int> s;
vector<int> v(n + );
for(int j = ; j <= n; j++)
scanf("%d", &v[j]);//读入要检验的序列。
int current = ;//指向输入的序列。
for(int j = ; j <= n; j++) {
s.push(j);
if(s.size() > m) break;
while(!s.empty() && s.top() == v[current]) {
s.pop();
current++;
}
}
if(current == n + ) flag = true;
if(flag) printf("YES\n");
else printf("NO\n");
}
return ;
}
//真的很厉害了,学习了。
1.使用一个current来指向当前的检验序列。
2.如果栈的大小已经大于了容量,那么就退出。
3.还有这个while循环是最关键的,只要top值等于当前current指向的,那么就弹出,并且指向下一个元素,非常厉害了。
//学习了,另一位大佬的代码思路也是相同的。
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