HDU 6375(双端队列 ~)

题意是有至多150000个双端队列，400000次简单操作，直接开会导致内存超限，所以用 STL 中的 map 和 deque ，而读入过大已经在题目中有所说明，直接用已经给出的快速读入即可。要注意的是在两个队列合并时，要用 insert 函数，直接一个一个操作会超时（自己对双端队列的 STL 还是不够熟悉......）代码如下：

 #include <bits/stdc++.h>
 using namespace std;
 const int N = ;
 map<int, deque<int> > q;
 void read(int &x){
     char ch = getchar();x = ;
     for (; ch < '' || ch > ''; ch = getchar());
     for (; ch >='' && ch <= ''; ch = getchar()) x = x *  + ch - '';
 }
 int main()
 {
     int a,n,m,u,v,w,val;
     while(~scanf("%d%d",&n,&m))
     {
         for(int i = ; i <= n; i++)
             q[i].clear();
         while(m--)
         {
             read(a);
             if(a == )
             {
                 read(u);
                 read(w);
                 read(val);
                 if(w == ) q[u].push_front(val);
                 else if(w == ) q[u].push_back(val);
             }
             else if(a == )
             {
                 read(u);
                 read(w);
                 if(q[u].empty())
                 {
                     puts("-1");
                     continue;
                 }
                 if(w == )
                 {
                     printf("%d\n",q[u].front());
                     q[u].pop_front();
                 }
                 else if(w == )
                 {
                     printf("%d\n",q[u].back());
                     q[u].pop_back();
                 }
             }
             else if(a == )
             {
                 read(u);
                 read(v);
                 read(w);
                 if(w == )
                 {
                     q[u].insert(q[u].end(),q[v].begin(),q[v].end());
                     q[v].clear();
                 }
                 else if(w == )
                 {
                     q[u].insert(q[u].end(),q[v].rbegin(),q[v].rend());
                     q[v].clear();
                 }
             }
         }

     }
     return ;
 }

一般来说 STL 用法简单，但是其中会带有许多与题目本身无关的功能，而用时也就会多出很多，用数组去模拟一些 STL 中的标准模板可以大幅减少不必要的耗时，本题中手写的双端队列用时不到用 STL 的一半，代码如下：

 #include<stdio.h>
 void read(int &x){
     char ch = getchar();x = ;
     for (; ch < '' || ch > ''; ch = getchar());
     for (; ch >='' && ch <= ''; ch = getchar()) x = x *  + ch - '';
 }
 struct record
 {
     int value;
     record *next;
     record *pre;
 }stack1[];
 int cnt;
 class listqueue
 {
 public:
     record *head,*tail;
     void pop()
     {
         if(!empty())
         {
             printf("%d\n",tail->value);
             tail = tail->pre;
             if(tail == )
                 head = ;
             else
                 tail->next = ;
         }
         else
             puts("-1");
     }
     bool empty()
     {
         return head == ;
     }
     void shift()
     {
         if(!empty())
         {
             printf("%d\n",head->value);
             head = head->next;
             if(head == )
                 tail = ;
             else
                 head->pre = ;
         }
         else
             puts("-1");
     }
     void clear()
     {
         head = tail = ;
     }
     void unshift(int n)
     {
         stack1[cnt].value = n;
         stack1[cnt].next = head;
         stack1[cnt].pre = ;
         if(head)
             head->pre = &stack1[cnt];
         head = &stack1[cnt];
         if(tail == )
             tail = head;
         cnt++;
     }
     void push(int n)
     {
         stack1[cnt].value = n;
         stack1[cnt].next = ;
         stack1[cnt].pre = tail;
         if(tail)
             tail->next = &stack1[cnt];
         tail = &stack1[cnt];
         if(head == )
             head = tail;
         cnt++;
     }
     void append(listqueue &v)
     {
         if(v.head)
         {
             if(head == )
                 head = v.head;
             else
             {
                 tail->next = v.head;
                 v.head->pre = tail;
             }
             tail = v.tail;
             v.clear();
         }
     }
     void reverse()
     {
         if(empty())
             return;
         record *temp,*temp1;
         for(temp = tail; temp != head; temp = temp->next)
         {
             temp1 = temp->next;
             temp->next = temp->pre;
             temp->pre = temp1;
         }
         temp1 = temp->next;
         temp->next = temp->pre;
         temp->pre = temp1;
         head = tail;
         tail = temp;
     }
 }rec[];
 int main()
 {
     int n,q,u,v,w,val;
     int type,i;
     while(~scanf("%d%d",&n,&q))
     {
         cnt = ;
         for(i = ; i <= n; i++)
         {
             rec[i].clear();
         }
         while(q--)
         {
             read(type);
             if(type == )
             {
                 read(u);
                 read(w);
                 read(val);
                 if(w == )
                     rec[u].unshift(val);
                 else
                     rec[u].push(val);
             }
             else if(type == )
             {
                 read(u);
                 read(w);
                 if(w == )
                     rec[u].shift();
                 else
                     rec[u].pop();
             }
             else
             {
                 read(u);
                 read(v);
                 read(w);
                 if(w == )
                     rec[v].reverse();
                 rec[u].append(rec[v]);
             }
         }
     }
     return ;
 }