1051 Pop Sequence (25 分)

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, ..., Nand 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

题⽬⼤意：有个容量限制为m的栈，分别把1，2，3，…，n⼊栈,

                  给出⼀个系列出栈顺序，问这些出栈 顺序是否可能～

分析：按照要求进⾏模拟。先把输⼊的序列接收进数组v。然后按顺序1~n把数字进栈，每进⼊⼀个数

字，判断有没有超过最⼤范围，超过了就break。如果没超过，设current = 1，从数组的第⼀个数字开

始，看看是否与栈顶元素相等，while相等就⼀直弹出栈，不相等就继续按顺序把数字压⼊栈～～最后

根据变量flag的bool值输出yes或者no～

 

#include 
#include 
#include 
using namespace std;
int main() {
 int m, n, k;
 scanf("%d %d %d", &m, &n, &k);
 for( int i = 0; i < k; i++) {
      bool flag = false;
      stack s;
      vector v(n + 1);
      for( int j = 1; j <= n; j++) 
	       scanf("%d", &v[j]);
      int  current = 1;
      for( int j = 1; j <= n; j++) {
           s.push(j);
      if(  s.size() > m ) break;
      while( !s.empty() && s.top() == v[current] ) {
             s.pop();
             current++;
      }
 }
 if( current == n + 1 ) 
      flag = true;
 if(  flag  ) 
      printf("YES\n");
 else printf("NO\n");
 }
 return 0;
}