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：
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int N = 150005; 4 map<int, deque<int> > q; 5 void read(int &x){ 6 char ch = getchar();x = 0; 7 for (; ch < '0' || ch > '9'; ch = getchar()); 8 for (; ch >='0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0'; 9 } 10 int main() 11 { 12 int a,n,m,u,v,w,val; 13 while(~scanf("%d%d",&n,&m)) 14 { 15 for(int i = 1; i <= n; i++) 16 q[i].clear(); 17 while(m--) 18 { 19 read(a); 20 if(a == 1) 21 { 22 read(u); 23 read(w); 24 read(val); 25 if(w == 0) q[u].push_front(val); 26 else if(w == 1) q[u].push_back(val); 27 } 28 else if(a == 2) 29 { 30 read(u); 31 read(w); 32 if(q[u].empty()) 33 { 34 puts("-1"); 35 continue; 36 } 37 if(w == 0) 38 { 39 printf("%d ",q[u].front()); 40 q[u].pop_front(); 41 } 42 else if(w == 1) 43 { 44 printf("%d ",q[u].back()); 45 q[u].pop_back(); 46 } 47 } 48 else if(a == 3) 49 { 50 read(u); 51 read(v); 52 read(w); 53 if(w == 0) 54 { 55 q[u].insert(q[u].end(),q[v].begin(),q[v].end()); 56 q[v].clear(); 57 } 58 else if(w == 1) 59 { 60 q[u].insert(q[u].end(),q[v].rbegin(),q[v].rend()); 61 q[v].clear(); 62 } 63 } 64 } 65 66 } 67 return 0; 68 }
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：
1 #include<stdio.h> 2 void read(int &x){ 3 char ch = getchar();x = 0; 4 for (; ch < '0' || ch > '9'; ch = getchar()); 5 for (; ch >='0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0'; 6 } 7 struct record 8 { 9 int value; 10 record *next; 11 record *pre; 12 }stack1[500010]; 13 int cnt; 14 class listqueue 15 { 16 public: 17 record *head,*tail; 18 void pop() 19 { 20 if(!empty()) 21 { 22 printf("%d ",tail->value); 23 tail = tail->pre; 24 if(tail == 0) 25 head = 0; 26 else 27 tail->next = 0; 28 } 29 else 30 puts("-1"); 31 } 32 bool empty() 33 { 34 return head == 0; 35 } 36 void shift() 37 { 38 if(!empty()) 39 { 40 printf("%d ",head->value); 41 head = head->next; 42 if(head == 0) 43 tail = 0; 44 else 45 head->pre = 0; 46 } 47 else 48 puts("-1"); 49 } 50 void clear() 51 { 52 head = tail = 0; 53 } 54 void unshift(int n) 55 { 56 stack1[cnt].value = n; 57 stack1[cnt].next = head; 58 stack1[cnt].pre = 0; 59 if(head) 60 head->pre = &stack1[cnt]; 61 head = &stack1[cnt]; 62 if(tail == 0) 63 tail = head; 64 cnt++; 65 } 66 void push(int n) 67 { 68 stack1[cnt].value = n; 69 stack1[cnt].next = 0; 70 stack1[cnt].pre = tail; 71 if(tail) 72 tail->next = &stack1[cnt]; 73 tail = &stack1[cnt]; 74 if(head == 0) 75 head = tail; 76 cnt++; 77 } 78 void append(listqueue &v) 79 { 80 if(v.head) 81 { 82 if(head == 0) 83 head = v.head; 84 else 85 { 86 tail->next = v.head; 87 v.head->pre = tail; 88 } 89 tail = v.tail; 90 v.clear(); 91 } 92 } 93 void reverse() 94 { 95 if(empty()) 96 return; 97 record *temp,*temp1; 98 for(temp = tail; temp != head; temp = temp->next) 99 { 100 temp1 = temp->next; 101 temp->next = temp->pre; 102 temp->pre = temp1; 103 } 104 temp1 = temp->next; 105 temp->next = temp->pre; 106 temp->pre = temp1; 107 head = tail; 108 tail = temp; 109 } 110 }rec[150100]; 111 int main() 112 { 113 int n,q,u,v,w,val; 114 int type,i; 115 while(~scanf("%d%d",&n,&q)) 116 { 117 cnt = 1; 118 for(i = 1; i <= n; i++) 119 { 120 rec[i].clear(); 121 } 122 while(q--) 123 { 124 read(type); 125 if(type == 1) 126 { 127 read(u); 128 read(w); 129 read(val); 130 if(w == 0) 131 rec[u].unshift(val); 132 else 133 rec[u].push(val); 134 } 135 else if(type == 2) 136 { 137 read(u); 138 read(w); 139 if(w == 0) 140 rec[u].shift(); 141 else 142 rec[u].pop(); 143 } 144 else 145 { 146 read(u); 147 read(v); 148 read(w); 149 if(w == 1) 150 rec[v].reverse(); 151 rec[u].append(rec[v]); 152 } 153 } 154 } 155 return 0; 156 }