02-线性结构4. Pop Sequence (25)

02-线性结构4. Pop Sequence (25)

时间限制
100 ms

内存限制
65536 kB

代码长度限制
8000 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 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

好几个月没写数据结构程序了，显得生疏，网查大侠方法，模拟整个进栈，检测，出栈的过程，这个问题就能慢慢解决了~

#include<iostream>
#include<stack>
using namespace std;

int main()
{
int M,N,K;
//	freopen("C:\\Users\\Jimko\\Desktop\\cin.txt","r",stdin);
cin>>M>>N>>K;
for(int i=0;i<K;i++){
stack<int> s;
bool flag = true;
int t=1;
for(int j=0;j<N;j++){		//测试数据逐个输入
int num;
cin>>num;
if(flag){
while(s.empty() || s.top() != num){
s.push(t);
if(s.size()>M){
flag = false;
break;
}
t++;			//有序进栈
}
if(flag && s.size()>=1 && s.top()==num)		//出栈条件
s.pop();
}
}
if(flag)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}