1051. Pop Sequence (25)
2017-05-21 08:39
393 查看
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
4000
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
参照算法笔记P259.
模拟,将1~n依次入栈,在入栈的过程中,如果入栈的元素恰好等于出栈序列 当前等待的元素,那么就让栈顶元素出栈,同时把出栈序列当前等待出栈的元素位置标记后移一位,。如果栈顶元素仍然等于出栈序列当前等待出栈的元素,则继续出栈。
#include<stdio.h>
#include<stack>
using namespace std;
int main(){
int m,n,k,i;
scanf("%d%d%d",&m,&n,&k);
int arr[1001];
while(k--){
stack<int>s;
for(i=0;i<n;i++){
scanf("%d",&arr[i]);
}
int now=0;
for(i=1;i<=n;i++){
s.push(i);
if(s.size()>m){
break;
}
while(!s.empty()&&s.top()==arr[now]){
s.pop();
now++;
}
}
if(s.empty()){
printf("YES\n");
}
else{
printf("NO\n");
}
}
}
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