1051. 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 51 2 3 4 5 6 73 2 1 7 5 6 47 6 5 4 3 2 15 6 4 3 7 2 11 7 6 5 4 3 2

Sample Output:

YESNONOYESNO
#include <stdio.h>#define MAXSIZE 1001int top;                    //指针，始终指向栈顶元素，若栈为空，则为-1 int main(){int S[MAXSIZE];   //定义数组用作栈 int Max,n,k,i,j,m,num,num_max,flag;scanf("%d%d%d",&Max,&n,&k);   //依次读取栈的长度，序列中元素的个数，序列的个数 for (i=0;i<k;i++)       {top=-1;              //指针，始终指向栈顶元素，若栈为空，则为-1 flag=0;              //开关，初始为0 num_max=0;           //一个序列中已经读取到的最大元素，初始化为0 for (j=0;j<n;j++){scanf("%d",&num); //读取一个元素 if (num>num_max)  //如果这个元素比已读取到最大元素还要大{for (m=num_max+1;m<=num;m++) //最大元素和这个元素之间的元素入栈 {S[++top]=m;if (top+1>Max) //如果入栈过程中超过栈的长度，则不存在这样的序列 {printf("NO\n");flag=1;break;}}num_max=num;      //重新将这个元素定义为最大元素 if (flag==1) break; top--;}else      //如果这个元素比已读取到最大元素小，则将其与栈顶元素相比，看其是不是栈顶元素 {if (S[top--]!=num)    {printf("NO\n");flag=1;break;}}}if (flag==0) printf("YES\n");   //如果一个序列的所有元素都符合规则，则输出YES while (j<n-1)     //如果循坏非正常结束，说明此序列不正确，把本序列剩下元素读取完，为下一个序列做准备 {scanf("%d",&num);j++;}   }return 0;}