UVa

题意

有一个长队包含几个小队, 每个小队连在一起排成一个长队, 有以下三种操作 :

① ENQUEUE 在队列中输入元素x
② DEQUEUE 输出长队队首并将其从队列中移除
③ STOP 结束处理

如果有队友就站到自己所属队列的最末, 如果没有队友就站到整个长队的最末

AC代码

#include <iostream>#include <cstdio>#include <map>#include <queue>#include <cstring>using namespace std;const int maxn = 1000 + 10;int main(){    int q,x;    int casenum = 0;    while( ~scanf("%d",&q) && q )    {        printf("Scenario #%d\n", ++casenum);        map<int, int> team;        for( int i = 1; i <= q; i++ )        {            int n;            scanf("%d",&n);            while( n-- ){                scanf("%d",&x);                team[x] = i;  //记录编号为x的人所在的队伍为第i队                //printf("%d,",team[x]);            }        }        queue<int> q, sq[maxn];        //q是长队的队列，sq[i]是小队i成员的队列        for(;;)        {            char s[maxn];            scanf("%s",s);            int mrk = s[0];            if( mrk == 'S' )            {                puts("");  //Print a blank lineafter each test case, even after the last one.                break;            }            else if( mrk == 'E' )            {                scanf("%d", &x);                int t = team[x];                if(sq[t].empty())                    q.push(t); //团队t进入队列                sq[t].push(x);            }            else if( mrk == 'D' )            {                int t = q.front();                printf("%d\n", sq[t].front());                sq[t].pop();                if(sq[t].empty())                    q.pop(); //团体t全体出队列            }        }    }    return 0;}