UVA 540 Team Queue

两个队列。把人分为t组，然后排队。当前到来的人要看之前有没有跟自己一组的人，如果有，就加到那个队的队尾，否则就加到大队的队尾。

两个队列，一个是大队，一个是同组人的队。has数组表示当前大队中有没有该组，p数组表示每个数对应的组。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<string>
#include<queue>
#include<cmath>
#include<map>
///LOOP
#define REP(i, n) for(int i = 0; i < n; i++)
#define FF(i, a, b) for(int i = a; i < b; i++)
#define FFF(i, a, b) for(int i = a; i <= b; i++)
#define FD(i, a, b) for(int i = a - 1; i >= b; i--)
#define FDD(i, a, b) for(int i = a; i >= b; i--)
///INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define RFI(n) scanf("%lf", &n)
#define RFII(n, m) scanf("%lf%lf", &n, &m)
#define RFIII(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define RFIV(n, m, k, p) scanf("%lf%lf%lf%lf", &n, &m, &k, &p)
#define RS(s) scanf("%s", s)
///OUTPUT
#define PN printf("\n")
#define PI(n) printf("%d\n", n)
#define PIS(n) printf("%d ", n)
#define PS(s) printf("%s\n", s)
#define PSS(s) printf("%s ", n)
///OTHER
#define PB(x) push_back(x)
#define CLR(a, b) memset(a, b, sizeof(a))
#define CPY(a, b) memcpy(a, b, sizeof(b))
#define display(A, n, m) {REP(i, n){REP(j, m)PIS(A[i][j]);PN;}}

using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int MOD = 100000000;
const int INFI = 1e9 * 2;
const LL LINFI = 1e17;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int N = 1111;
const int M = 1111111;
const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};

queue<int> q1, q2
;
int p[M];
bool has
;
char s[11];

int main()
{
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);

int t, k, a, cas = 1;
while(RI(t), t)
{
CLR(p, 0);
CLR(has, 0);
while(!q1.empty())q1.pop();
REP(i, t + 1)while(!q2[i].empty())q2[i].pop();
REP(i, t)
{
RI(k);
REP(j, k)
{
RI(a);
p[a] = i + 1;
}
}
printf("Scenario #%d\n", cas++);
while(RS(s), s[0] != 'S')
{
if(s[0] == 'E')
{
RI(a);
k = p[a];
q2[k].push(a);
if(!has[k])
{
q1.push(k);
has[k] = 1;
}
}
else
{
k = q1.front();
a = q2[k].front();
q2[k].pop();
if(q2[k].empty())q1.pop(), has[k] = 0;
PI(a);
}
}
puts("");
}
return 0;
}