1051. Pop Sequence (25)

1051. Pop Sequence (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

---

提交代码

 

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<queue>
 6 #include<vector>
 7 #include<cmath>
 8 #include<string>
 9 #include<map>
10 #include<set>
11 #include<stack>
12 using namespace std;
13 int line[1005];
14 int main(){
15     //freopen("D:\\INPUT.txt","r",stdin);
16     int n,m,k;
17     scanf("%d %d %d",&m,&n,&k);
18     int i,j,l,maxnum;
19     for(i=0;i<k;i++){
20         stack<int> s;//注意初始化
21         maxnum=0;
22         for(j=0;j<n;j++){
23             scanf("%d",&line[j]);
24         }
25         for(j=0;j<n;j++){
26             while(line[j]>maxnum){
27                 s.push(++maxnum);
28             }
29             if(s.size()>m){
30                 break;
31             }
32             if(line[j]<=maxnum&&s.top()==line[j]){
33                 s.pop();
34             }
35             else{
36                 break;
37             }
38         }
39         if(j==n){
40             printf("YES");
41         }
42         else{
43             printf("NO");
44         }
45         printf("\n");
46     }
47     return 0;
48 }

 

转载于:https://www.cnblogs.com/Deribs4/p/4772144.html