题目

首先是开闭区间的处理，我们把\(1.5\)这种数加进来，用\([1.5,6]\)来表示\((2,6]\)

根据离散数学的基本知识，尝试把五个操作转化成人话

1. 把\([x,y]\)变成\(1\)

2. 把\([0,x-1]\)和\([y+1,inf]\)变成\(0\)

3. 把\([x,y]\)变成\(0\)

4. 把\([x,y]\)取反，之后来一个二操作

5. 把\([x,y]\)取反

于是线段树维护一下区间覆盖就好了

记得覆盖标记和取反标记只能同时存在一个就好了

代码

#include<algorithm>

#include<iostream>

#include<cstring>

#include<cstdio>

#define max(a,b) ((a)>(b)?(a):(b))

#define min(a,b) ((a)<(b)?(a):(b))

#define LL long long

#define re register

#define maxn 150005

inline int read() {

    int x=0,f=0;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();}

    if(ch==')')f=1;

    return x*2-f;

}

char opt[3];

int rev[maxn<<2],tag[maxn<<2];

int d[maxn<<2],l[maxn<<2],r[maxn<<2];

void build(int x,int y,int i) {

    l[i]=x,r[i]=y;tag[i]=-1;

    if(x==y) return;

    int mid=x+y>>1;

    build(x,mid,i<<1),build(mid+1,y,i<<1|1);

}

inline void pushdown(int i) {

    if(tag[i]!=-1) {

        rev[i<<1]=rev[i<<1|1]=0;

        tag[i<<1]=tag[i<<1|1]=tag[i];

        d[i<<1]=d[i<<1|1]=tag[i];

        tag[i]=-1;

    }

    if(rev[i]) {

        d[i<<1|1]^=1,d[i<<1]^=1;

        if(tag[i<<1]!=-1) tag[i<<1]^=1;

            else rev[i<<1]^=1;

        if(tag[i<<1|1]!=-1) tag[i<<1|1]^=1;

            else rev[i<<1|1]^=1;

        rev[i]=0;

    }

}

void change(int x,int y,int val,int i) {

    if(x<=l[i]&&y>=r[i]) {

        d[i]=val;tag[i]=val;

        if(rev[i]) rev[i]=0;

        return;

    }

    pushdown(i);

    int mid=l[i]+r[i]>>1;

    if(x<=mid) change(x,y,val,i<<1);

    if(y>=mid+1) change(x,y,val,i<<1|1);

}

void solve(int x,int y,int i) {

    if(x<=l[i]&&y>=r[i]) {

        d[i]^=1;

        if(tag[i]!=-1) tag[i]^=1;

            else rev[i]^=1;

        return;

    }

    pushdown(i);

    int mid=l[i]+r[i]>>1;

    if(x<=mid) solve(x,y,i<<1);

    if(y>=mid+1) solve(x,y,i<<1|1);

}

int query(int pos,int i) {

    if(l[i]==r[i]) return d[i];

    pushdown(i);

    int mid=l[i]+r[i]>>1;

    if(pos<=mid) return query(pos,i<<1);

    return query(pos,i<<1|1);

}

int main() {

    //freopen("a.in","r",stdin);

    //freopen("a.out","w",stdout);

    build(1,150000,1);

    while(scanf("%s",opt)!=EOF) {

        int x=read()+2,y=read()+2;

        if(opt[0]=='U') change(x,y,1,1);

        if(opt[0]=='I') change(1,x-1,0,1),change(y+1,150000,0,1);

        if(opt[0]=='D') change(x,y,0,1);

        if(opt[0]=='C') change(1,x-1,0,1),change(y+1,150000,0,1),solve(x,y,1);

        if(opt[0]=='S') solve(x,y,1);

        //if(opt[0]=='I') break;

    }

    int st=-1,lst=-1,flag=0;

    for(re int i=1;i<=150000;i++) {

        if(query(i,1))  {

            if(st==-1) st=i;

            lst=i;

        }

        else {

            if(st!=-1) {

                if(flag) putchar(' ');

                    else flag=1;

                if(st&1) putchar('(');

                    else putchar('[');

                printf("%d",st/2-1);

                putchar(',');

                printf("%d",(lst+1)/2-1);

                if(lst&1) putchar(')');

                    else putchar(']');

            }

            lst=st=-1;

        }

    }

    if(!flag) puts("empty set");

    return 0;

}