P3968 [TJOI2014]电源插排

P3968 [TJOI2014]电源插排

线段树维护最长空区间及左端点位置，这个和$nlongn$的动态最大子序和差不多，就不多解释了

$n$较大哈希优化空间

My complete code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
using namespace std;
typedef long long LL;
const LL maxn=3000000;
const LL hs=299987;
inline LL Read(){
    LL x=0,f=1; char c=getchar();
    while(c<'0'||c>'9'){
        if(c=='-') f=-1; c=getchar();
    }
    while(c>='0'&&c<='9'){
        x=(x<<3)+(x<<1)+c-'0'; c=getchar();
    }return x*f;
}
struct node{
    LL son[2],lx,rx,mx,mxi,col;
}tree[maxn];
LL n,q,nod=1,root=1;
LL que[300000][2];
inline LL Get(LL x){
    LL u=x%hs+1;
    while(que[u][0]&&que[u][0]!=x)
        u=(u+1)%n;
    return u;
}
inline void Update(node &now,node son0,node son1,LL l,LL r,LL mid){
    if(!son0.col)
        son0.lx=son0.rx=son0.mx=(mid-l+1);
    if(!son1.col)
        son1.lx=son1.rx=son1.mx=(r-mid);
    if(son0.mx>son1.mx)
        now.mx=son0.mx,now.mxi=son0.mxi;
    else
        now.mx=son1.mx,now.mxi=son1.mxi;
    if(son0.rx+son1.lx>now.mx||(son0.rx+son1.lx==now.mx&&mid-son0.rx+1>now.mxi))
        now.mx=son0.rx+son1.lx,
        now.mxi=mid-son0.rx+1;
    now.lx=son0.lx,now.rx=son1.rx;
    if(!son0.col)
        now.lx+=son1.lx;
    if(!son1.col)
        now.rx+=son0.rx;
    now.col=son0.col+son1.col;
}
void Change(LL &now,LL l,LL r,LL x,LL c){
    if(!now)
        now=++nod;
    if(l==r){
        tree[now].col=c;
        tree[now].lx=tree[now].rx=tree[now].mx=!c;
        if(!c)
            tree[now].mxi=l;
        return;
    }
    LL mid=(l+r)>>1;
    x<=mid?Change(tree[now].son[0],l,mid,x,c):Change(tree[now].son[1],mid+1,r,x,c);
    Update(tree[now],tree[tree[now].son[0]],tree[tree[now].son[1]],l,r,mid);
}
LL Query(LL now,LL l,LL r,LL lt,LL rt){
    if(!now)
        return 0;
    if(lt<=l&&rt>=r)
        return tree[now].col;
    LL mid=(l+r)>>1;
    LL sum=0;
    if(lt<=mid)
        sum+=Query(tree[now].son[0],l,mid,lt,rt);
    if(rt>mid)
        sum+=Query(tree[now].son[1],mid+1,r,lt,rt);
    return sum;
}
int main(){
    n=Read(),q=Read();
    tree[root].mx=n,
    tree[root].mxi=1;
    while(q--){
        LL op=Read();
        if(!op){
            LL l=Read(),r=Read();
            printf("%lld
",Query(root,1,n,l,r));
        }else{
            LL x=Get(op);
            if(que[x][1]){
                Change(root,1,n,que[x][1],0);
                que[x][1]=0;
            }else{
                que[x][0]=op;
                LL now=(tree[1].mxi*2+tree[1].mx)>>1;
                Change(root,1,n,now,1);
                que[x][1]=now;
            }
        }
    }
    return 0;
}