UVA 540 Team Queue

题目翻译摘自《算法禁赛入门经典》

题目大意

　　有 t 个团队的人正在排一个长队。每次新来一个人时，如果他有队友在排队，那么这个新人会插队到最后一个队友的身后。如果没有任何一个队友排队，则他会排到长队的队尾。输入每个团队中所有队员的编号，要求支持如下3种指令（前两种指令可以穿插进行）。

ENQUEUEx：编号为 x 的人进入长队。
DEQUEUE：长队的队首出队。
STOP：停止模拟。

　　对于每个DEQUEUE指令，输出出队的人的编号。

分析

　　嵌套队列。

代码如下

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3  
  4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  5 #define Rep(i,n) for (int i = 0; i < (n); ++i)
  6 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
  7 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
  8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
  9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 12  
 13 #define pr(x) cout << #x << " = " << x << "  "
 14 #define prln(x) cout << #x << " = " << x << endl
 15  
 16 #define LOWBIT(x) ((x)&(-x))
 17  
 18 #define ALL(x) x.begin(),x.end()
 19 #define INS(x) inserter(x,x.begin())
 20  
 21 #define ms0(a) memset(a,0,sizeof(a))
 22 #define msI(a) memset(a,inf,sizeof(a))
 23 #define msM(a) memset(a,-1,sizeof(a))
 24 
 25 #define MP make_pair
 26 #define PB push_back
 27 #define ft first
 28 #define sd second
 29  
 30 template<typename T1, typename T2>
 31 istream &operator>>(istream &in, pair<T1, T2> &p) {
 32     in >> p.first >> p.second;
 33     return in;
 34 }
 35  
 36 template<typename T>
 37 istream &operator>>(istream &in, vector<T> &v) {
 38     for (auto &x: v)
 39         in >> x;
 40     return in;
 41 }
 42  
 43 template<typename T1, typename T2>
 44 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
 45     out << "[" << p.first << ", " << p.second << "]" << "
";
 46     return out;
 47 }
 48 
 49 inline int gc(){
 50     static const int BUF = 1e7;
 51     static char buf[BUF], *bg = buf + BUF, *ed = bg;
 52     
 53     if(bg == ed) fread(bg = buf, 1, BUF, stdin);
 54     return *bg++;
 55 } 
 56 
 57 inline int ri(){
 58     int x = 0, f = 1, c = gc();
 59     for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
 60     for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
 61     return x*f;
 62 }
 63 
 64 template<class T>
 65 inline string toString(T x) {
 66     ostringstream sout;
 67     sout << x;
 68     return sout.str();
 69 }
 70  
 71 typedef long long LL;
 72 typedef unsigned long long uLL;
 73 typedef pair< double, double > PDD;
 74 typedef pair< int, int > PII;
 75 typedef pair< int, PII > PIPII;
 76 typedef pair< string, int > PSI;
 77 typedef pair< int, PSI > PIPSI;
 78 typedef set< int > SI;
 79 typedef set< PII > SPII;
 80 typedef vector< int > VI;
 81 typedef vector< VI > VVI;
 82 typedef vector< SI > VSI;
 83 typedef vector< PII > VPII;
 84 typedef map< int, int > MII;
 85 typedef map< int, string > MIS;
 86 typedef map< int, PII > MIPII;
 87 typedef map< PII, int > MPIII;
 88 typedef map< string, int > MSI;
 89 typedef map< string, string > MSS;
 90 typedef map< PII, string > MPIIS;
 91 typedef map< PII, PII > MPIIPII;
 92 typedef multimap< int, int > MMII;
 93 typedef multimap< string, int > MMSI;
 94 //typedef unordered_map< int, int > uMII;
 95 typedef pair< LL, LL > PLL;
 96 typedef vector< LL > VL;
 97 typedef vector< VL > VVL;
 98 typedef priority_queue< int > PQIMax;
 99 typedef priority_queue< int, VI, greater< int > > PQIMin;
100 const double EPS = 1e-10;
101 const LL inf = 0x7fffffff;
102 const LL infLL = 0x7fffffffffffffffLL;
103 const LL mod = 1e9 + 7;
104 const int maxN = 1e3 + 7;
105 const LL ONE = 1;
106 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
107 const LL oddBits = 0x5555555555555555;
108 
109 int T, t;
110 string op;
111 
112 int main(){
113     //freopen("MyOutput.txt","w",stdout);
114     //freopen("input.txt","r",stdin);
115     //INIT();
116     while(cin >> t) {
117         if(!t) break;
118         MII mii; // stu_id->team_id
119         queue< int > Q;
120         queue< int > team[t + 1];
121         int vis[t + 1]; // 标识某队伍在不在大队列中 
122         ms0(vis);
123         
124         For(i, 1, t) {
125             int x, y;
126             cin >> x;
127             Rep(j, x) {
128                 cin >> y;
129                 mii[y] = i;
130             }
131         }
132         
133         printf("Scenario #%d
", ++T);
134         while(cin >> op) {
135             if(op == "STOP") break;
136             if(op[0] == 'E') {
137                 int x, y;
138                 cin >> x;
139                 y = mii[x];
140                 team[y].push(x);
141                 if(!vis[y]) {
142                     vis[y] = 1;
143                     Q.push(y);
144                 }
145             } 
146             else {
147                 int x = Q.front();
148                 cout << team[x].front() << endl;
149                 team[x].pop();
150                 if(team[x].empty()) {
151                     vis[x] = 0;
152                     Q.pop();
153                 }
154             }
155         }
156         cout << endl;
157     }
158     return 0;
159 }

View Code

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原文地址：https://www.cnblogs.com/zaq19970105/p/11051926.html