（线段树）poj3225-Help with Intervals

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

Input

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

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.

Sample Input

U [1,5]
D [3,3]
S [2,4]
C (1,5)
I (2,3]

Sample Output

(2,3)

#include <iostream>
//#include<bits/stdc++.h>
#include <stack>
#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int MAX=65535;
struct node
{
    int st,rev;
}a[500*MAX];
bool vi[3*MAX];

void xxor(int k)
{
    if(a[k].st!=-1)//如果区间状态一致，直接改变区间的情况
        a[k].st^=1;
    else
        a[k].rev^=1;//不然标记区间0，1互换1次
}
void pushdown(int k)
{
    if(a[k].st!=-1)//如果区间状态一致
    {
        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
        a[2*k].rev=a[2*k+1].rev=0;
        a[k].st=-1;//并且将父结点状态改回
    }
    if(a[k].rev)//如果需要0，1互换
    {
        xxor(2*k);
        xxor(2*k+1);
        a[k].rev=0;
    }
}
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
{
    if(L>=l&&R<=r)//如果当前区间已经包含在了目标区间之中
    {
        if(operation=='U')//如果取并集
        {
            a[k].st=1;//T区间所有元素值都赋1
            a[k].rev=0;//0，1互换情况初始化为0
        }
        else if(operation=='D')//如果取差集，差掉T
        {
            a[k].st=0;//T区间所有元素都赋0
            a[k].rev=0;//0，1互换情况初始化为0
        }
        else if(operation=='C'||operation=='S')//如果是原集合被差掉或异或集
        {
            xxor(k);
        }
        return;
    }
//    if(L+1==R)
//        return;
    pushdown(k);
    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
    {
        update(operation,l,r,L,(L+R)/2,2*k);
    }
    else//若不然，当前区间左半部分已经在目标区间之外
    {
        if(operation=='I'||operation=='C')
        {
            a[2*k].st=0;a[2*k].rev=0;
        }
    }
    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
    {
        update(operation,l,r,(L+R)/2+1,R,2*k+1);
    }
    else
    {
        if(operation=='I'||operation=='C')
        {
            a[2*k+1].st=a[2*k+1].rev=0;
        }
    }
}
void query(int l,int r,int k)
{
    if(a[k].st==1)//如果这个区间完整的都是状态1
    {
        for(int i=l;i<=r;i++)
            vi[i]=true;
        return;
    }
    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
        return;
    if(l==r)
        return;
    pushdown(k);
    query(l,(l+r)/2,2*k);
    query((l+r)/2+1,r,2*k+1);
}
int main()
{
//    int n;
//    cin>>n;
//    scanf("%d
",&n);
    char op,ls,rs,tem;//tem接收逗号
    int l,r;
//    memset(vi,false,sizeof(vi));
    a[1].st=a[1].rev=0;
//    while(--n)
//    for(int q=1;q<=n;q++)

//        scanf("%c %c%d,%d%c
",&op,&ls,&l,&r,&rs);
        while(~scanf("%c %c%d,%d%c
",&op,&ls,&l,&r,&rs))
//        cin>>op>>ls>>l>>tem>>r>>rs;
      {
//          scanf("%c %c%d,%d%c
",&op,&ls,&l,&r,&rs);
        l*=2;
        r*=2;
        if(ls=='(')
            l++;
        if(rs==')')
            r--;
        update(op,l,r,0,2*MAX,1);
//        printf("%d
",n);
    }
//    printf("~~!
");
    query(0,2*MAX,1);
//    printf("!!
");
    int st=-1,en;//输出时的区间
    bool flag=false;
    for(int i=0;i<=2*MAX;i++)
    {
        if(vi[i])
        {
            if(st==-1)
                st=i;
            en=i;
        }
        else
        {
            if(st!=-1)
            {
                if(flag)
                    printf(" ");
//                if(st%2==0)
//                    printf("[");
//                else
//                    printf("(");
                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
//                if(en%2==0)
//                    printf("]");
//                else
//                    printf(")");
                flag=true;
                st=-1;
            }

        }
    }
    if(!flag)
        printf("empty set");
    return 0;
}