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
1 #include<stdio.h>
2 //本解法主要思路为逆向放置
3 //即将测试用例中的数从最后一个开始放置
4 //如果能在一定容量的堆栈下输出N~1
5 //将放与取操作取反
6 //即可满足题意
7 int main(){
8 int M,N,K;
9 int target;
10 int top;
11 int data[1001];
12 scanf("%d %d %d",&M,&N,&K);
13 int b[N];
14 int n ;
15 while(K--){
16 target = N;
17 top = -1;
18 n = N;
19 while(n){
20 scanf("%d",&b[N-n]);
21 n--;
22 }
23 n = N-1;
24 while(n+1){
25 //printf("%d\n",top);
26 top++;
27 if(top >= M){
28 break;
29 }
30 data[top] = b[n];
31
32 while(data[top] == target&&top >-1){
33 target--;
34 //printf("%d",top);
35 top--;
36 // printf("%d",top);
37 }
38 n--;
39 }
40
41 //printf("%d",top);
42 //printf("%d",target);
43 if(top == -1){
44 printf("YES");
45 }else{
46 printf("NO");
47 }
48 if(K != 0){
49 printf("\n");
50 }
51 }
52 return 0;
53 }
