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

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<queue>
#include<vector>
#include<cmath>
#include<string>
#include<map>
#include<set>
#include<stack>
using namespace std;
int line[1005];
int main(){
    //freopen("D:\INPUT.txt","r",stdin);
    int n,m,k;
    scanf("%d %d %d",&m,&n,&k);
    int i,j,l,maxnum;
    for(i=0;i<k;i++){
        stack<int> s;//注意初始化
        maxnum=0;
        for(j=0;j<n;j++){
            scanf("%d",&line[j]);
        }
        for(j=0;j<n;j++){
            while(line[j]>maxnum){
                s.push(++maxnum);
            }
            if(s.size()>m){
                break;
            }
            if(line[j]<=maxnum&&s.top()==line[j]){
                s.pop();
            }
            else{
                break;
            }
        }
        if(j==n){
            printf("YES");
        }
        else{
            printf("NO");
        }
        printf("
");
    }
    return 0;
}