1051. Pop Sequence (25)

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

#include<cstdio>
#include<stack>
using namespace std;
const int maxn = 1010;
int arr[maxn];
stack<int> st;

int main(){
    int m,n,k;
    scanf("%d%d%d",&m,&n,&k);
    while(k--){ //清空栈中元素 
        while(!st.empty()){
            st.pop();
        }
        int i;
        for(i = 1; i <= n; i++){
            scanf("%d",&arr[i]);  //输入待匹配的序列 
        }
        int current = 1;
        bool flag = true;
        for(i = 1; i <= n;i++){
            st.push(i);
            if(st.size() > m){
                flag = false;
                break;
            }
            while(!st.empty() && st.top() == arr[current]){ //反复弹出 要用while 
                st.pop();
                current++;
            }
        }
        if(st.empty() == true&&flag == true) printf("YES\n");
        else printf("NO\n");
    }

    return 0;
}