HDU1702-----队列简单题

最新推荐文章于 2019-11-25 17:13:58 发布

原创最新推荐文章于 2019-11-25 17:13:58 发布 · 965 阅读

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本文深入探讨了AI音视频处理领域中的关键技术，特别是视频分割与语义识别，详细解释了如何利用这些技术实现更智能、更高效的音视频处理流程。通过实际案例分析，展示在不同场景下应用这些技术所能带来的显著效果。

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Problem Description

ACboy was kidnapped!!
he miss his mother very much and is very scare now.You can't image how dark the room he was put into is, so poor :(.
As a smart ACMer, you want to get ACboy out of the monster's labyrinth.But when you arrive at the gate of the maze, the monste say :" I have heard that you are very clever, but if can't solve my problems, you will die with ACboy."
The problems of the monster is shown on the wall:
Each problem's first line is a integer N(the number of commands), and a word "FIFO" or "FILO".(you are very happy because you know "FIFO" stands for "First In First Out", and "FILO" means "First In Last Out").
and the following N lines, each line is "IN M" or "OUT", (M represent a integer).
and the answer of a problem is a passowrd of a door, so if you want to rescue ACboy, answer the problem carefully!

Input

The input contains multiple test cases.
The first line has one integer,represent the number oftest cases.
And the input of each subproblem are described above.

Output

For each command "OUT", you should output a integer depend on the word is "FIFO" or "FILO", or a word "None" if you don't have any integer.

Sample Input

4
4 FIFO
IN 1
IN 2
OUT
OUT
4 FILO
IN 1
IN 2
OUT
OUT
5 FIFO
IN 1
IN 2
OUT
OUT
OUT
5 FILO
IN 1
IN 2
OUT
IN 3
OUT



FIFO先进先出---队列，FILO先进后出---栈








#include<cstdio>
#include<stack>
#include<queue>
#include<algorithm>
using namespace std;
int main(){
	int a, b, c;
	char s[10];
	scanf("%d", &a);
	while(a--){
		scanf("%d %s", &b, s);
		if(s[2] == 'F'){
			queue <int> que;//定义名为que的队列
			for(int i = 0; i < b; i++){
				scanf("%s", s);
				if(s[0] == 'I'){
					scanf("%d", &c);
					que.push(c);//放入队列
				}
				else{
					if(que.empty()){//空队列
						printf("None\n");
					}
					else{
						printf("%d\n", que.front());//输出首队列
						que.pop();//清除首队列
					}
				}
			}
		}
		else{
			stack <int> ata;//定义名为ata的栈
			for(int i = 0; i < b; i++){
				scanf("%s", s);
				if(s[0] == 'I'){
					scanf("%d", &c);
					ata.push(c);//放入栈
				}
				else{
					if(ata.empty()){//空栈
						printf("None\n");
					}
					else{
						printf("%d\n", ata.top());//输出首栈
						ata.pop();//清除首栈
					}
				}
			}
		}
	}
	return 0;
}