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
理解要点：1.要pop n，前提是要push 1 2 3 … n-1。 num为既定序列1 2 3 … 例如，input为 4 3 2 1的4时，首先要push 1 2 3 才能push 4 pop 4 出栈。
2.stack.size() > M(the maximum capacity of the stack) 超出栈容量不可能。
思路：当栈顶元素不是input 则push（num++）直到input，随后pop input；或者stack.size() > M(the maximum capacity of the stack) 检测出不可能。
#include<iostream>
#include<stack>
using namespace std; int main()
{
int M; //maximum capacity of the stack
int N; //the length of push sequence
int K; //the number of pop sequence to be checked
cin >> M >> N >> K;
bool flag = true;
int input, num; //num= 1.2.3.4.5... input为依次输入的检测序列值
stack<int> sta; for(int i = ; i < K; i++) {
num = ;
flag = true;
for(int j = ; j < N; j++) {
cin >> input;
while( sta.size() <= M && num ) {
if(sta.empty() || sta.top() != input) {
sta.push(num ++);
}else if(sta.top() == input) {
sta.pop();
break;
}
}
if(sta.size() > M ) //超出栈容量
flag = false;
}
if(flag)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl; while(!sta.empty()) //清空栈
sta.pop();
}
return ;
}
参考：http://www.2cto.com/kf/201311/254409.html 开始误入了比大小的误区，这个比较细致易理解。记一笔。