Pat(Advanced Level)Practice--1051(Pop Sequence)

Pat1051代码

题目描述：

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

AC代码：

#include<cstdio>#include<stack>using namespace std;int main(int argc,char *argv[]){int m,n,k;scanf("%d%d%d",&m,&n,&k);while(k--){int flag=1,cur=1;int i,temp;stack<int> s;for(i=1;i<=n;i++){scanf("%d",&temp);if(flag){while(s.empty()||s.top()!=temp){s.push(cur);if(s.size()>m){flag=0;break;}cur++;}if(flag&&s.top()==temp)s.pop();}}printf("%s\n",flag?"YES":"NO");}return 0;}

如果栈顶元素与输入元素不相等，则一直压入元素，如果超出stack的大小则停止。

如果相等则弹出栈顶的元素。

如此进行下去....