HDU-1387-Team Queue

HDU-1387-Team Queue

大家好,又见面了,我是全栈君,今天给大家准备了Idea注册码。

Team Queue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1259    Accepted Submission(s): 430




Problem Description
Queues and Priority Queues are data structures which are known to most computer scientists. The Team Queue, however, is not so well known, though it occurs often in everyday life. At lunch time the queue in front of the Mensa is a team queue, for example. 

In a team queue each element belongs to a team. If an element enters the queue, it first searches the queue from head to tail to check if some of its teammates (elements of the same team) are already in the queue. If yes, it enters the queue right behind them. If not, it enters the queue at the tail and becomes the new last element (bad luck). Dequeuing is done like in normal queues: elements are processed from head to tail in the order they appear in the team queue. 

Your task is to write a program that simulates such a team queue. 

 


Input
The input will contain one or more test cases. Each test case begins with the number of teams t (1<=t<=1000). Then t team descriptions follow, each one consisting of the number of elements belonging to the team and the elements themselves. Elements are integers in the range 0 – 999999. A team may consist of up to 1000 elements. 

Finally, a list of commands follows. There are three different kinds of commands: 

ENQUEUE x – enter element x into the team queue 

DEQUEUE – process the first element and remove it from the queue 

STOP – end of test case 

The input will be terminated by a value of 0 for t. 

 


Output
For each test case, first print a line saying “Scenario #k”, where k is the number of the test case. Then, for each DEQUEUE command, print the element which is dequeued on a single line. Print a blank line after each test case, even after the last one. 

 


Sample Input
   
   
2 3 101 102 103 3 201 202 203 ENQUEUE 101 ENQUEUE 201 ENQUEUE 102 ENQUEUE 202 ENQUEUE 103 ENQUEUE 203 DEQUEUE DEQUEUE DEQUEUE DEQUEUE DEQUEUE DEQUEUE STOP 2 5 259001 259002 259003 259004 259005 6 260001 260002 260003 260004 260005 260006 ENQUEUE 259001 ENQUEUE 260001 ENQUEUE 259002 ENQUEUE 259003 ENQUEUE 259004 ENQUEUE 259005 DEQUEUE DEQUEUE ENQUEUE 260002 ENQUEUE 260003 DEQUEUE DEQUEUE DEQUEUE DEQUEUE STOP 0

 


Sample Output
   
   
Scenario #1 101 102 103 201 202 203 Scenario #2 259001 259002 259003 259004 259005 260001

 


Source
 


这几天都在做栈和队列的题。刚刚起步。想多做点!

这个是一个模拟队列的题!

我之前用的链表写的。由于非常久没用链表了。可是这题用链表非常麻烦,确实。纠结我好久,可是还是写出来了,代码不是非常清晰,提交一遍然后TLE了,唉:-(


然后网上看了看别人的代码,发现自己还有非常多不足之处!

!所以说新手还是得多看看别人的代码啊!那个代码思路还是非常清晰的,先贴在这里,以后再来回顾回顾!



我的TLE代码(链表:没用c++里的queue写,一是想练习一下链表。二是想模拟那个过程):


#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#include <queue>
#include <stack>
using namespace std;

typedef struct que
{
	int da;
	struct que* next;
}*queu, node;

struct fun
{
	int da;
	int te;
}te_me[1000000];

int me_num;

int Len(queu q)
{
	int len=0;
	while(q->next)
	{
		q = q->next;
		len++;
	}
	return len;
}

int query(int a)
{
	for(int i=0; i<me_num; i++)
		if(a == te_me[i].da)return te_me[i].te; 
}

int Entry_q(int a, queu q, int len)
{
	int team = query(a);
	for(int i=0; i < len; i++)
	{
		if(team == query(q->da)  && team != query(q->next->da))
		{
			queu p = (node*)malloc(sizeof(node)); p->next=NULL;
			p->da = a;
			p->next = q->next;
			q->next = p;
			return 1;
		}
		else q = q->next;
	}
	return 0;
}

int main()
{
	int T, count=0;
	while(scanf("%d", &T), T)
	{
		count++;
		int n; me_num=0;
		for(int k = 1; k <= T; k++)
		{
			scanf("%d", &n); int m;
			while(n--)
			{
				scanf("%d", &m);
				te_me[me_num].da = m;
				te_me[me_num].te = k;
				me_num++;
			}
		}
		printf("Scenario #%d\n", count);
		queu q = (node*)malloc(sizeof(node)); q->next=NULL;
		queu front=q, rear=q, p;
		char fun[10]; int elem;
		while(scanf("%s", fun)!=EOF)
		{
			if(!strcmp(fun, "ENQUEUE")){
				scanf("%d", &elem);
				if(front == rear){
					rear->da = elem;
					p = (node*)malloc(sizeof(node)); p->next=NULL;
					rear->next = p; rear = p;
				}
				else if(!Entry_q(elem, front, Len(front))) {
					rear->da = elem;
					p = (node*)malloc(sizeof(node)); p->next=NULL;
					rear->next = p; rear = p; 
				}
			}
			else if(!strcmp(fun, "DEQUEUE")){
				if(front->da)printf("%d\n", front->da);
				queu fe=front;
				front = fe->next;
				free(fe);
			}
			else if(!strcmp(fun, "STOP")){
				break;
			}
		}
		printf("\n");
	}
	return 0;
} 


AC代码(93ms):


#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

#define	MAX_RANK 1000000
#define MAX_QUE  1000
#define MAX_N    1000
#define CMD_CHAR 30

int team[MAX_RANK];
queue<int> que[MAX_QUE];
queue<int> bigQue;
void init();
int main()
{
	int cases = 1;

	int teamM;
	while (scanf("%d", &teamM) == 1 && teamM) {
		// init
		init();

		// enter team
		int n;
		memset(team, 0, sizeof(team));
		for (int team_NO = 0; scanf("%d", &n) == 1; team_NO++) {
			for (int i = 0; i < n; i++) {
				int num;
				scanf("%d%*c", &num);
				team[num] = team_NO;
			}
		}

		// read commands
		printf("Scenario #%d\n", cases++);
		while (true) {
			char cmd[CMD_CHAR];
			scanf("%s", cmd);

			if (strcmp(cmd, "ENQUEUE") == 0) {
				int num;	
				scanf("%d%*c", &num);

				if (que[team[num]].empty()) {
					bigQue.push(team[num]);
				}
				que[team[num]].push(num);
			} else if (strcmp(cmd, "DEQUEUE") == 0) {
				int whitch_team = bigQue.front();
				printf("%d\n", que[whitch_team].front());
				que[whitch_team].pop();
				if (que[whitch_team].empty()) {
					bigQue.pop();
				}
			} else {
				printf("\n");
				break;
			}
		}
		
	}

	return 0;
}
void init()
{
	while (!bigQue.empty()) {
		bigQue.pop();
	}

	for (int i = 0; i < MAX_QUE; i++) {
		while (!que[i].empty()) {
			que[i].pop();
		}
	}
}

版权声明:本文博客原创文章,博客,未经同意,不得转载。

版权声明:本文内容由互联网用户自发贡献,该文观点仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 举报,一经查实,本站将立刻删除。

发布者:全栈程序员-用户IM,转载请注明出处:https://javaforall.cn/117438.html原文链接:https://javaforall.cn

【正版授权,激活自己账号】: Jetbrains全家桶Ide使用,1年售后保障,每天仅需1毛

【官方授权 正版激活】: 官方授权 正版激活 支持Jetbrains家族下所有IDE 使用个人JB账号...

(0)


相关推荐

  • 概率论中 PDF,PMF,CDF的含义[通俗易懂]

    概率论中 PDF,PMF,CDF的含义[通俗易懂]概率论中PDF,PMF,CDF的含义在概率论中,我们经常能碰到这样几个概念PDF,PMF,CDF,这里就简单介绍一下PDF:概率密度函数(probabilitydensityfunction),在数学中,连续型随机变量的概率密度函数(在不至于混淆时可以简称为密度函数)是一个描述这个随机变量的输出值,在某个确定的取值点附近的可能性的函数。概率密度函数都是针对连续性随机变量的,对于连续性随机变量,都是针对某一段区间的取值,在一个点的取值都是几乎为0的,所以我们研究连续性随机变量时,都是取变量在一段

  • css-day05笔记-清除浮动&学成网布局准备工作

    css-day05笔记-清除浮动&学成网布局准备工作typora-copy-images-to:media第01阶段.WEB基础:css-day05笔记-清除浮动&学成网布局准备工作一.清除浮动1.为什么要清除浮动因为父级盒子很多情况下,不方便给高度,但是子盒子浮动就不占有位置,最后父级盒子高度为0,就影响了后面的标准流盒子。总结:由于浮动元素不再占用原文档流的位置,所以它会对后面的元素排版产生影响准确地说…

  • goland破解激活码【在线破解激活】

    goland破解激活码【在线破解激活】,https://javaforall.cn/100143.html。详细ieda激活码不妨到全栈程序员必看教程网一起来了解一下吧!

  • 求生之路2机枪mod_求生之路2好看的枪械mod名字

    求生之路2机枪mod_求生之路2好看的枪械mod名字动态子弹(弹匣)数量修改工具:GCFScape、notepad++、vpk.exe自己在搜索引擎输入文件名找资源教程:教程均假设修改smg的静态子弹和动态子弹,其他枪械修改一样,只不过修改对象不同。1.先找出武器原始的数据文件(1).打开GCFScape,左上角File-open。找到“\Steam\SteamApps\common\Left4Dead2\left4dead…

    2022年10月29日
  • sprintf()函数的用法总结

    sprintf()函数的用法总结sprintf()函数的程序用例:#include#includeintmain(void){charbuffer[80];sprintf(buffer,”AnapproximationofPiis%f\n”,M_PI);puts(buffer);return0;}sprintf的作用是将一个格式化的字符串输出到一个

  • Java继承

    Java继承一:继承的概述1.继承的定义继承:就是子类继承父类的属性和行为,使得子类对象具有与父类相同的属性、相同的行为。子类可以直接访问父类中的非私有的属性和行为。–注:父类又称为超类或者基类。子类又称为派生类!2.继承的格式通过 extends 关键字,可以声明一个子类继承另外一个父类,定义格式如下:class父类{…}class子类extends父类{…}二、关于继承之后的成员变量1.当成员变量不重名如果子类父类中出现不重名的成员变量,这时的访问是没有影

发表回复

您的电子邮箱地址不会被公开。

关注全栈程序员社区公众号