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.
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， 请联系我们举报，一经查实，本站将立刻删除。
发布者：全栈程序员-站长，转载请注明出处：https://javaforall.net/117438.html原文链接：https://javaforall.net
