UVa 540:team QUEUE Data Structure topic

Topic Link:

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=103&page=show_ problem&problem=481

Type of topic: data structure, two fork tree

Sample input:

2
3 102
3 201 203 ENQUEUE The ENQUEUE
201
ENQUEUE 102 ENQUEUE ENQUEUE >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
enque UE 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
102
201
203

Scenario #2
259001
259002 259003
259004
259005
260001

The main effect of the topic:

There's a so-called team sort, giving out T teams and members of these teams. and then line up.

ENQUEUE x: Inserts an X into the queue. If the queue does not have a member of that element, the last face of the queue is inserted. If there is already a member of his team in the queue, then insert the last player.

Dequeue: Remove the elements of the team head and output

Stop: Stopping

Ideas for solving problems:

This article URL address: http://www.bianceng.cn/Programming/sjjg/201410/45535.htm

If you are familiar with the linked list, it is not difficult to simulate this process. The list in STL is very useful, is a two-way circular chain list. The end () member function returns the iteration of the first link. A key process for a topic is which team to query element x (can be accelerated with a binary lookup, improve the speed is very objective), you can also open a large array, with coordinate mapping element values, the array to save the team number, so that the search elements of the team only need O (1) time, is the fastest. A more critical step, then, is that every time you insert a search, you'll be tle if you have to find it once for each insertion. You can also open an iterator array to hold the iterator of the last position in the queue for a team.

Code:

#include <iostream> #include <cstdio> #include <cstring> #include <list> #include &LT;STRING&G  
T  
    
#include <algorithm> using namespace std;  
int T, Commandnum,posteam;  
int team[1005][1005];  
    
Char command[10];  
list<int>m_list;  
List<int>::iterator lastit[1005];  int ele[1000000]; Used to map which team the element belongs to.   
Subscript corresponds to the element value//Find out which troop this number is in.  
    int searchteam (int num) {int i,j; For (i=0 i<t; ++i) {//Two point lookup if (Binary_search (team[i]+1, team[i]+1+team[i][0, num)) r  
    Eturn i; //If not found, return-1.   
In fact, the topic to the element must belong to a certain team, is I want to complicate the return-1;  
    } void Input () {int i,j;  
        For (i=0 i<t; ++i) {scanf ("%d", &team[i][0]);  
            For (j=1 j<=team[i][0]; ++j) {scanf ("%d", &team[i][j]);  
        ELE[TEAM[I][J]] = i;  
        } Lastit[i] = M_list.end ();    To sort, prepare for two points in sort (team[i]+1, team[i]+1+team[i][0]);
    }//queue void Enqueue (int num) {posteam=searchteam (num);    
    Use the following method to get the team faster//posteam = Ele[num]; if (lastit[posteam] = = M_list.end ()) {//inserted in M_list.end (), the effect is the same as push_back lastit[posteam] = M_list.inser   
    T (Lastit[posteam], num);  
        } else{++lastit[posteam];  
           
    Lastit[posteam] = M_list.insert (Lastit[posteam], num);  
    } void Dequeue () {if (M_list.empty ()) return;   
    printf ("%d\n", M_list.front ());  
    List<int>::iterator it = Lastit[m_list.front ()];  
            If the pop-up element is the last of the remaining teams in the queue, modify the for (int i=0; i<t; ++i) {if (Lastit[i]==m_list.begin ()) {  
            Lastit[i] = M_list.end ();  
        Break  
} m_list.pop_front ();  
    } void Solve () {if (!m_list.empty ()) m_list.clear ();  
        while (~SCANF ("%s", command)) {if (!strcmp (command, "STOP")) break; if (!strcmp command, "EnqueUE ")) {scanf ("%d ", &commandnum);  
        Enqueue (Commandnum);  
        else if (!strcmp (command, "Dequeue")) {dequeue ();  
    ()} int main () {freopen ("Input.txt", "R", stdin);  
    int Cas=1;  
        while (~SCANF ("%d", &t) && t) {Input ();  
        printf ("Scenario #%d\n", cas++);  
        Solve ();  
    printf ("\ n");  
return 0; }