（deque）度度熊学队列

Sample Input

2 10
1 1 1 23
1 1 0 233
2 1 1 
1 2 1 2333
1 2 1 23333
3 1 2 1
2 2 0
2 1 1
2 1 0
2 1 1

Sample Output

23
-1
2333
233
23333

提示

由于读入过大，C/C++ 选手建议使用读入优化。

一个简单的例子：

void read(int &x){
	char ch = getchar();x = 0;
	for (; ch < '0' || ch > '9'; ch = getchar());
	for (; ch >='0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
}

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<functional>
#include<deque>
using namespace std;
typedef long long ll;
int n,q1;

ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}


int main(){
    ll u,v,w,val,op;
    while(~scanf("%d%d",&n,&q1)){
       deque<ll> q[n+1];
       while(q1--){
           op=read();
           if(op==1){
                u=read();
                w=read();
                val=read();
                if(w==0){
                    q[u].push_front(val);
                }else q[u].push_back(val);
           }else if(op==2){
                u=read();
                w=read();
                if(q[u].empty()) printf("-1\n");
                else{
                    if(w==0){
                        printf("%lld\n",q[u].front());
                        q[u].pop_front();
                    }else{
                        printf("%lld\n",q[u].back());
                        q[u].pop_back();
                    }
                }
           }else{
                u=read();
                v=read();
                w=read();
                if(w==0){
                    while(!q[v].empty()){
                        q[u].push_back(q[v].front());
                        q[v].pop_front();
                    }
                }else{
                    while(!q[v].empty()){
                        q[u].push_back(q[v].back());
                        q[v].pop_back();
                    }
                }
           }
       }
    }
    return 0;
}