【题解】Luogu P2572 [SCOI2010]序列操作

原题传送门：P2572 [SCOI2010]序列操作

这题好弱智啊

裸的珂朵莉树

前置芝士：珂朵莉树

窝博客里对珂朵莉树的介绍

没什么好说的自己看看吧

操作1：把区间内所有数推平成0，珂朵莉树基本操作

操作2：把区间内所有数推平成1，珂朵莉树基本操作

操作3：把区间内所有数取反（异或1），split后扫描一遍，值域取反

操作4：区间和，珂朵莉树基本操作

操作5：区间最长连续的1，暴力扫描累加

就是这样简单

好像跑的比线段树还快？？？喵喵喵

#pragma GCC optimize("O3")
#include <bits/stdc++.h>
#define IT set<node>::iterator
using namespace std;
inline int read()
{
    register int 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*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
inline int Max(register int a,register int b)
{
    return a>b?a:b;
}
struct node
{
    int l,r;
    mutable bool v;
    node(int L, int R=-1, bool V=0):l(L), r(R), v(V) {}
    bool operator<(const node& o) const
    {
        return l < o.l;
    }
};
set<node> s;
inline IT split(register int pos)
{
    IT it = s.lower_bound(node(pos));
    if (it != s.end() && it->l == pos)
        return it;
    --it;
    int 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 assign_val(register int l,register int r,register bool val)
{
    IT itr = split(r+1),itl = split(l);
    s.erase(itl, itr);
    s.insert(node(l, r, val));
}
inline void rev(register int l,register int r)
{
    IT itr = split(r+1),itl = split(l);
    for(; itl != itr; ++itl)
        itl->v ^= 1;
}
inline int sum(register int l,register int r)
{
    IT itr = split(r+1),itl = split(l);
    int res = 0;
    for (; itl != itr; ++itl)
        res += itl->v ? itl->r - itl->l + 1 : 0;
    return res;
}
inline int count(register int l,register int r)
{
    int res=0,temp=0;
    IT itr = split(r+1),itl = split(l);
    for(; itl != itr; ++itl)
    {
        if(itl->v == false)
        {
            res = Max(res, temp);
            temp=0;
        }
        else
            temp += itl->r - itl->l + 1;
    }
    return Max(res, temp);
}
int main()
{
    int n=read(),m=read();
    for(register int i=0;i<n;++i)
        s.insert(node(i,i,read()));
    s.insert(node(n,n,0));
    while(m--)
    {
        int op=read(),a=read(),b=read();
        if(op==0)
            assign_val(a,b,0);
        else if(op==1)
            assign_val(a,b,1);
        else if(op==2)
            rev(a,b);
        else if(op==3)
            printf("%d\n",sum(a,b));
        else
            printf("%d\n",count(a,b));
    }
    return 0;
 } 

原文地址：https://www.cnblogs.com/yzhang-rp-inf/p/9763257.html