【bzoj1251】序列终结者

用魔法平衡树的实现！

反正没人会看的。

#include<bits/stdc++.h>
#define rat 5
#define newnode(s,v,a,b) (&(*st[cnt++]=Node(s,v,a,b)))
#define up(cur) pushdown(cur),pushup(cur),pushup(rt->lc),pushup(rt)
#define N 400005
using namespace std;
struct Node{
    int size,val,rev,addv;
    Node *lc,*rc;
    Node(int s,int v,Node *a,Node *b):size(s),val(v),lc(a),rc(b),rev(0),addv(0){}
    Node(){}
}*nul,*rt,*st[N],t[N];
int n,m,cnt;
inline void pushdown(Node *o){
    register Node *l=o->lc,*r=o->rc;
    if(o->addv){
        if(!l->size)o->val+=o->addv;
        else l->addv+=o->addv,r->addv+=o->addv;
        o->addv=0;
    }
    if(o->rev){
        if(l->size){
            l->rev^=1;r->rev^=1;
            swap(o->lc,o->rc);
        }o->rev=0;
    }
}
inline void pushup(Node *o){
    register Node *l=o->lc,*r=o->rc;
    if(l->size)o->size=l->size+r->size,o->val=max(l->val+l->addv,r->val+r->addv);
}
inline Node* merge(Node *a,Node *b){
    pushdown(a);pushdown(b);pushup(a);pushup(b);
    return newnode(a->size+b->size,max(a->val,b->val),a,b);
}
void split(int x,Node *o){
    pushdown(o);
    if(x>o->lc->size){
        split(x-o->lc->size,o->rc);o->lc=merge(o->lc,o->rc->lc);
        st[--cnt]=o->rc;o->rc=o->rc->rc;
    }
    else if(x<o->lc->size){
        split(x,o->lc);o->rc=merge(o->lc->rc,o->rc);
        st[--cnt]=o->lc;o->lc=o->lc->lc;
    }
    pushup(o);
}
Node* build(int l,int r){
    if(l==r)return newnode(1,0,nul,nul);
    int mid=(l+r)>>1;
    return merge(build(l,mid),build(mid+1,r));
}
inline Node* cut(int l,int r){
    split(r,rt);split(l-1,rt->lc);return rt->lc->rc;
}
inline int read(){
    int f=1,x=0;char ch;
    do{ch=getchar();if(ch==‘-‘)f=-1;}while(ch<‘0‘||ch>‘9‘);
    do{x=x*10+ch-‘0‘;ch=getchar();}while(ch>=‘0‘&&ch<=‘9‘);
    return f*x;
}
int main(){
    n=read();m=read();
    for(int i=0;i<=N;i++)st[i]=&t[i];
    nul=new Node(0,0,0,0);
    rt=build(0,n+1);
    while(m--){
        int opt=read(),l=read()+1,r=read()+1;
        Node* o=cut(l,r);
        if(opt==1){o->addv+=read();up(o);}
        if(opt==2){o->rev^=1;up(o);}
        if(opt==3){pushdown(o);pushup(o);printf("%d\n",o->val);}
    }
}