【题解】Luogu CF817F MEX Queries

原题传送门

817，我突然想到了某8位质数

这题珂以说是珂朵莉树的模板

三个操作都肥肠简单，前两个区间赋值，第三个区间0变1,1变0

每次输出从头开始扫描就行（我忘了珂朵莉树的性质，竟然还动态维护最左边0的位置）

#include <bits/stdc++.h> 
#define getchar nc
#define ll long long
#define IT set<node>::iterator
using namespace std;
inline char nc(){
    static char buf[100000],*p1=buf,*p2=buf; 
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++; 
}
inline ll read()
{
    register ll x=0,f=1;register char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
    return x*f;
}
inline void write(register ll x)
{
    if(!x)putchar('0');if(x<0)x=-x,putchar('-');
    static int sta[25];register int tot=0;
    while(x)sta[tot++]=x%10,x/=10;
    while(tot)putchar(sta[--tot]+48);
}
inline ll Min(register ll a,register ll b)
{
    return a<b?a:b;
}
struct node
{
    ll l,r;
    mutable bool v;
    node(ll L, ll R=-1, bool V=0):l(L), r(R), v(V) {}
    bool operator<(const node& o) const
    {
        return l < o.l;
    }
};
set<node> s;
ll ans=1;
inline IT split(register ll pos)
{
    IT it = s.lower_bound(node(pos));
    if (it != s.end() && it->l == pos) 
        return it;
    --it;
    ll L = it->l, R = it->r;
    bool V = it->v;
    s.erase(it);
    s.insert(node(L, pos-1, V));
    return s.insert(node(pos, R, V)).first;
}
inline void findnext()
{
    IT it = split(ans);
    for(;;++it)
        if(!it->v)
        {
            ans=it->l;
            return;
        }
}
inline void assign_val(register ll l,register ll r,register bool val)
{
    IT itr = split(r+1),itl = split(l);
    s.erase(itl, itr);
    s.insert(node(l, r, val));
    if(val&&l<=ans&&ans<=r)
        findnext();
    else if(!val)
        ans=Min(ans,l);
}
inline void reverse(register ll l,register ll r)
{
    IT itr = split(r+1),itl = split(l);
    bool f=true;
    if(l<=ans&&ans<=r)
    {
        for(; itl != itr; ++itl)
        {
            itl->v^=1;
            if(f&&!itl->v)
            {
                f=false;
                ans=itl->l;
            }
        }
        if(f)
            findnext();
    }
    else
        for(; itl != itr; ++itl)
        {
            itl->v^=1;
            if(f&&!itl->v)
            {
                f=false;
                ans=Min(ans,itl->l);
            }
        }       
}
int main()
{
    int q=read();
    s.insert(node(1,1e19,0));
    while(q--)
    {
        int opt=read();
        ll l=read(),r=read();
        if(opt==1)
            assign_val(l,r,1);
        else if(opt==2)
            assign_val(l,r,0);
        else
            reverse(l,r);
        write(ans),puts("");
    }
    return 0;
}

转载于:https://www.cnblogs.com/yzhang-rp-inf/p/10350935.html