LogLoader, Inc. is a company specialized in providing products for analyzing logs. While Ikki is working on graduation design, he is also engaged in an internship at LogLoader. Among his tasks, one is to write a module for manipulating time intervals, which have confused him a lot. Now he badly needs your help.

In discrete mathematics, you have studied several basic set operations, namely union, intersection, relative complementation and symmetric difference, which naturally apply to the specialization of sets as intervals.. For your quick reference they are summarized in the table below:

Operation	Notation	Definition
Union	A ∪ B	{x : x ∈ A or x ∈ B}
Intersection	A ∩ B	{x : x ∈ A and x ∈ B}
Relative complementation	A ? B	{x : x ∈ A but B}
Symmetric difference	A ⊕ B	(A ? B) ∪ (B ? A)

Ikki has abstracted the interval operations emerging from his job as a tiny programming language. He wants you to implement an interpreter for him. The language maintains a set S, which starts out empty and is modified as specified by the following commands:

Command	Semantics
U T	S ← S ∪ T
I T	S ← S ∩ T
D T	S ← S ? T
C T	S ← T ? S
S T	S ← S ⊕ T

The input contains exactly one test case, which consists of between 0 and 65,535 (inclusive) commands of the language. Each command occupies a single line and appears like

X T

where X is one of ‘U’, ‘I’, ‘D’, ‘C’ and ‘S’ and T is an interval in one of the forms (a,b), (a,b], [a,b) and [a,b] (a, b ∈ Z, 0 ≤ a ≤ b ≤ 65,535), which take their usual meanings. The commands are executed in the order they appear in the input.

End of file (EOF) indicates the end of input.

Output the set S as it is after the last command is executed as the union of a minimal collection of disjoint intervals. The intervals should be printed on one line separated by single spaces and appear in increasing order of their endpoints. If S is empty, just print “empty set” and nothing else.

  1 #include <iostream>
  2 //#include<bits/stdc++.h>
  3 #include <stack>
  4 #include <queue>
  5 #include <cstdio>
  6 #include <cstring>
  7 #include <algorithm>
  8 using namespace std;
  9 typedef long long ll;
 10 typedef unsigned long long ull;
 11 const int MAX=65535;
 12 struct node
 13 {
 14     int st,rev;
 15 }a[500*MAX];
 16 bool vi[3*MAX];
 17 
 18 void xxor(int k)
 19 {
 20     if(a[k].st!=-1)//如果区间状态一致，直接改变区间的情况
 21         a[k].st^=1;
 22     else
 23         a[k].rev^=1;//不然标记区间0，1互换1次
 24 }
 25 void pushdown(int k)
 26 {
 27     if(a[k].st!=-1)//如果区间状态一致
 28     {
 29         a[2*k].st=a[2*k+1].st=a[k].st;//将父结点的状态传递到子节点
 30         a[2*k].rev=a[2*k+1].rev=0;
 31         a[k].st=-1;//并且将父结点状态改回
 32     }
 33     if(a[k].rev)//如果需要0，1互换
 34     {
 35         xxor(2*k);
 36         xxor(2*k+1);
 37         a[k].rev=0;
 38     }
 39 }
 40 void update(char operation,int l,int r,int L,int R,int k)//[l,r]是目标区间，[L,R]是此时区间
 41 {
 42     if(L>=l&&R<=r)//如果当前区间已经包含在了目标区间之中
 43     {
 44         if(operation==‘U‘)//如果取并集
 45         {
 46             a[k].st=1;//T区间所有元素值都赋1
 47             a[k].rev=0;//0，1互换情况初始化为0
 48         }
 49         else if(operation==‘D‘)//如果取差集，差掉T
 50         {
 51             a[k].st=0;//T区间所有元素都赋0
 52             a[k].rev=0;//0，1互换情况初始化为0
 53         }
 54         else if(operation==‘C‘||operation==‘S‘)//如果是原集合被差掉或异或集
 55         {
 56             xxor(k);
 57         }
 58         return;
 59     }
 60 //    if(L+1==R)
 61 //        return;
 62     pushdown(k);
 63     if(l<=(L+R)/2)//如果当前区间左半部分存在有部分在目标区间内的可能
 64     {
 65         update(operation,l,r,L,(L+R)/2,2*k);
 66     }
 67     else//若不然，当前区间左半部分已经在目标区间之外
 68     {
 69         if(operation==‘I‘||operation==‘C‘)
 70         {
 71             a[2*k].st=0;a[2*k].rev=0;
 72         }
 73     }
 74     if(r>(L+R)/2)//如果当前区间的右半部分存在有部分在目标区间内的可能
 75     {
 76         update(operation,l,r,(L+R)/2+1,R,2*k+1);
 77     }
 78     else
 79     {
 80         if(operation==‘I‘||operation==‘C‘)
 81         {
 82             a[2*k+1].st=a[2*k+1].rev=0;
 83         }
 84     }
 85 }
 86 void query(int l,int r,int k)
 87 {
 88     if(a[k].st==1)//如果这个区间完整的都是状态1
 89     {
 90         for(int i=l;i<=r;i++)
 91             vi[i]=true;
 92         return;
 93     }
 94     else if(a[k].st==0)//若此区间状态是已经完整的标明为状态0的，就不用再往下看了
 95         return;
 96     if(l==r)
 97         return;
 98     pushdown(k);
 99     query(l,(l+r)/2,2*k);
100     query((l+r)/2+1,r,2*k+1);
101 }
102 int main()
103 {
104 //    int n;
105 //    cin>>n;
106 //    scanf("%d\n",&n);
107     char op,ls,rs,tem;//tem接收逗号
108     int l,r;
109 //    memset(vi,false,sizeof(vi));
110     a[1].st=a[1].rev=0;
111 //    while(--n)
112 //    for(int q=1;q<=n;q++)
113 
114 //        scanf("%c %c%d,%d%c\n",&op,&ls,&l,&r,&rs);
115         while(~scanf("%c %c%d,%d%c\n",&op,&ls,&l,&r,&rs))
116 //        cin>>op>>ls>>l>>tem>>r>>rs;
117       {
118 //          scanf("%c %c%d,%d%c\n",&op,&ls,&l,&r,&rs);
119         l*=2;
120         r*=2;
121         if(ls==‘(‘)
122             l++;
123         if(rs==‘)‘)
124             r--;
125         update(op,l,r,0,2*MAX,1);
126 //        printf("%d\n",n);
127     }
128 //    printf("~~!\n");
129     query(0,2*MAX,1);
130 //    printf("!!\n");
131     int st=-1,en;//输出时的区间
132     bool flag=false;
133     for(int i=0;i<=2*MAX;i++)
134     {
135         if(vi[i])
136         {
137             if(st==-1)
138                 st=i;
139             en=i;
140         }
141         else
142         {
143             if(st!=-1)
144             {
145                 if(flag)
146                     printf(" ");
147 //                if(st%2==0)
148 //                    printf("[");
149 //                else
150 //                    printf("(");
151                 printf("%c%d,%d%c",st&1?‘(‘:‘[‘,st/2,(en+1)/2,en&1?‘)‘:‘]‘);//因为区间右侧做的是--，故需要+1再除二
152 //                if(en%2==0)
153 //                    printf("]");
154 //                else
155 //                    printf(")");
156                 flag=true;
157                 st=-1;
158             }
159 
160         }
161     }
162     if(!flag)
163         printf("empty set");
164     return 0;
165 }