poj-2259-Team Queue【优先队列】

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Team Queue

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 4459 Accepted: 1544

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. 
Warning: A test case may contain up to 200000 (two hundred thousand) commands, so the implementation of the team queue should be efficient: both enqueing and dequeuing of an element should only take constant time.

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

23 101 102 1033 201 202 203ENQUEUE 101ENQUEUE 201ENQUEUE 102ENQUEUE 202ENQUEUE 103ENQUEUE 203DEQUEUEDEQUEUEDEQUEUEDEQUEUEDEQUEUEDEQUEUESTOP25 259001 259002 259003 259004 2590056 260001 260002 260003 260004 260005 260006ENQUEUE 259001ENQUEUE 260001ENQUEUE 259002ENQUEUE 259003ENQUEUE 259004ENQUEUE 259005DEQUEUEDEQUEUEENQUEUE 260002ENQUEUE 260003DEQUEUEDEQUEUEDEQUEUEDEQUEUESTOP0

Sample Output

Scenario #1101102103201202203Scenario #2259001259002259003259004259005260001

思路：杭电上的数据太弱，不用优先队列也可以过的 

优先队列：

#include<cstdio>#include<algorithm>#include<cstring>#include<queue>using namespace std;int fa[1000010];int cnt[1010]; // 表示编号为 i的队列里面的元素个数 int mark[1010];  struct node{int id; // 队列编号 int rank; // 所有队列的所有元素个数 int ele;bool operator < (node x) const // 成员函数实现重载（一个参数） {  // 从小到大 if(id!=x.id)return id>x.id;return rank>x.rank;}/*bool friend operator < (node x,node y) // 友元函数实现重载 （两个参数） {if(x.id!=y.id)return x.id>y.id;return x.rank>y.rank;}*/}team;int main(){int t,text=0;while(scanf("%d",&t),t){memset(cnt,0,sizeof(cnt));memset(mark,0,sizeof(mark));for(int i=1;i<=t;i++){int n;scanf("%d",&n);for(int j=1;j<=n;j++){int x;scanf("%d",&x);fa[x]=i;}}priority_queue<node> Q;printf("Scenario #%d\n",++text);int d=1,k=1;char str[20];while(scanf("%s",str),str[0]!='S'){if(str[0]=='E'){int x;scanf("%d",&x);if(cnt[fa[x]]){cnt[fa[x]]++;team.id=mark[fa[x]];team.ele=x;team.rank=d++;Q.push(team);}else{cnt[fa[x]]++;mark[fa[x]]=k;team.id=k++;team.ele=x;team.rank=d++;Q.push(team);}}else{node ans=Q.top();Q.pop();printf("%d\n",ans.ele);cnt[fa[ans.ele]]--;}}puts("");}return 0;}

杭电上也能过的水码：

#include<cstdio>#include<algorithm>#include<cstring>#include<queue>#include<map>using namespace std;bool vis[1010];map<int,int> M;queue<int> Q,team[1010];int n;int main(){    int t,text=0;    while(scanf("%d",&t),t)    {        M.clear();        while(!Q.empty())    Q.pop();        for(int i=1;i<=n;i++)        {            while(!team[i].empty())                team[i].pop();        }        memset(vis,0,sizeof(vis));        for(int que=1;que<=t;que++)        {            scanf("%d",&n);            for(int i=1;i<=n;i++)            {                int ele;                scanf("%d",&ele);                M[ele]=que;            }        }        printf("Scenario #%d\n",++text);        char str[20];        while(scanf("%s",str),str[0]!='S')        {            if(str[0]=='E')            {                int x;                scanf("%d",&x);                int id=M[x];                team[id].push(x);                if(!vis[id])                {                    Q.push(id);                    vis[id]=1;                }            }            if(str[0]=='D')            {                int k=Q.front();                printf("%d\n",team[k].front());                team[k].pop();                if(team[k].empty())                {                    Q.pop();                    vis[k]=0;                    }            }            }        puts("");    }    return 0;}