PAT 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

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这道题就是考察对stack的操作，代码如下：

#include <iostream>#include <stack>#include <cstring>using namespace std;int main(){int M,N,K;cin>>M>>N>>K;int result[1001];stack<int> s;while(K--){memset(result,0,sizeof(result));while(!s.empty())s.pop();for(int i=1;i<=N;i++)cin>>result[i];int j=1,flag=1,i;for(i=1;i<=N;i++){if(s.size()==M)s.pop();s.push(i);while(result[j]<=i&&result[j]>0){int f=s.top();if(f==result[j]){s.pop();j++;}else{flag=0;break;}}if(flag==0)break;}if(i<=N)cout<<"NO"<<endl;elsecout<<"YES"<<endl;}return 0;}