Pop Sequence

Pop Sequence

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

AC Code

#include<stdio.h>
//本解法主要思路为逆向放置
//即将测试用例中的数从最后一个开始放置
//如果能在一定容量的堆栈下输出N~1
//将放与取操作取反
//即可满足题意
int main(){
  int M,N,K;
  int target;
  int top;
  int data[1001];
  scanf("%d %d %d",&M,&N,&K);
  int b[N];
  int n ;
  while(K--){
    target = N;
    top = -1;
    n = N;
    while(n){
      scanf("%d",&b[N-n]);
      n--;
    }
    n = N-1;
    while(n+1){
        //printf("%d
",top);
      top++;
      if(top >= M){
        break;
      }
      data[top] = b[n];
       
      while(data[top] == target&&top >-1){
        target--;
         //printf("%d",top);
        top--;
       // printf("%d",top); 
      }
      n--;
    }
    
       //printf("%d",top);
       //printf("%d",target);
    if(top == -1){
      printf("YES");
    }else{
      printf("NO");
    }
    if(K != 0){
      printf("
");
    }
  }
  return 0;
}