PAT 1051 Pop Sequence (25 分)

返回

1051 Pop Sequence (25 分)

 

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
using namespace std;
typedef long long ll;
int m,n,k;
int a[1005] = {0};

int main(){
    cin >> m >> n >> k;
    while(k--){
        int flag = 1;
        memset(a,0,sizeof(a));
        for(int i=0;i < n;i++){
            cin >> a[i];
        }
        for(int i=0;i < n;i++){
            int cnt = 0;
            int fflag = 1;
            int temp = 1005;
            for(int j=i+1;j < n;j++){
                if(a[i] > a[j]){
                    cnt++;
                    if(a[j] > temp) fflag = 0;
                    temp = a[j];
                } 
            }
            if(cnt > (m-1) || fflag == 0){
                cout << "NO" << endl;
                flag = 0;
                break;
            }
        }
        if(flag) cout << "YES" << endl;
    }
    
    return 0;
} 

合法出栈顺序首先得每个数后面不能有多于m-1的数比他小，不然装不下。

其次就是后面比他小的数一定要按递减排列

转载于:https://www.cnblogs.com/cunyusup/p/11460935.html