洛谷P3178 树上操作 [HAOI2015] 树链剖分

正解:树链剖分+线段树

解题报告:

传送门!

树链剖分+线段树算是基操了趴,,,

就无脑码码码,没有任何含金量,不需要动脑子,然后码量其实也不大,就很爽

比树剖的板子还要板子一些hhhhh

放下代码就欧克了QwQ

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define int long long
#define gc getchar()
#define ls(x) (x<<1)
#define rs(x) ((x<<1)|1)
#define t(i) edge[i].to
#define w(i) edge[i].wei
#define fy(i) edge[i].fy
#define ri register int
#define rb register bool
#define rc register char
#define rp(i,x,y) for(ri i=x;i<=y;++i)
#define my(i,x,y) for(ri i=x;i>=y;--i)
#define e(i,x) for(ri i=head[x];i;i=edge[i].nxt)
 
+;
int n,q,ed_cnt,head[N],fa[N],top[N],sz[N],sn[N],dfn[N],low[N],rk[N],dfn_cnt,val[N];
struct ed{int to,nxt;}edge[N];
struct node{int dat,tag;}tr[N];
 
il int read()
{
    rc ch=gc;ri x=;rb y=;
    '))ch=gc;
    ;
    )+(x<<)+(ch^'),ch=gc;
    return y?x:-x;
}
il char rd(){rc ch=gc;while(ch!='C' && ch!='Q')ch=gc;return ch;}
il void ad(ri x,ri y){edge[++ed_cnt]=(ed){x,head[y]};head[y]=ed_cnt;edge[++ed_cnt]=(ed){y,head[x]};head[x]=ed_cnt;}
;fa[x]=fat;e(i,x)if(t(i)^fat){dfs1(t(i),x),sz[x]+=sz[t(i)];if(sz[t(i)]>sz[sn[x]] || !sz[x])sn[x]=t(i);}}
void dfs2(ri x,ri tp)
{
    top[x]=tp;rk[dfn[x]=low[x]=++dfn_cnt]=x;if(sn[x])dfs2(sn[x],tp);low[x]=max(low[x],low[sn[x]]);
    e(i,x)if(t(i)^fa[x] && t(i)^sn[x])dfs2(t(i),t(i)),low[x]=low[t(i)];
}
il void pushdown(ri x,ri l,ri r)
{
    if(tr[x].tag)
    {
        ri mid=(l+r)>>;
        tr[ls(x)].tag+=tr[x].tag;tr[ls(x)].dat+=tr[x].tag*(mid-l+);
        tr[rs(x)].tag+=tr[x].tag;tr[rs(x)].dat+=tr[x].tag*(r-mid);
        tr[x].tag=;
    }
}
il void pushup(ri x){tr[x].dat=tr[ls(x)].dat+tr[rs(x)].dat;}
void build(ri x,ri l,ri r)
{
    if(l==r){tr[x].dat=val[rk[l]];return;}
    ri mid=(l+r)>>;
    build(ls(x),l,mid);build(rs(x),mid+,r);
    pushup(x);
}
void modify(ri x,ri l,ri r,ri to_l,ri to_r,ri dat)
{
    );return;}
    pushdown(x,l,r);
    ri mid=(l+r)>>;
    if(mid>=to_l)modify(ls(x),l,mid,to_l,to_r,dat);
    ,r,to_l,to_r,dat);
    pushup(x);
}
int query(ri x,ri l,ri r,ri to_l,ri to_r)
{
    ;
    if(to_l<=l && r<=to_r)return tr[x].dat;
    pushdown(x,l,r);
    ri mid=(l+r)>>,ret=;
    if(mid>=to_l)ret+=query(ls(x),l,mid,to_l,to_r);
    ,r,to_l,to_r);
    return ret;
}
il ,,n,dfn[x],dfn[x],dat);}
il ,,n,dfn[x],low[x],dat);}
il ;,,n,dfn[top[x]],dfn[x]),x=fa[top[x]];return ret;}
 
main()
{
//    freopen("3178.in","r",stdin);freopen("3178.out","w",stdout);
    n=read();q=read();rp(i,,n)val[i]=read();rp(i,,n-){ri x=read(),y=read();ad(x,y);}dfs1(,);dfs2(,);build(,,n);
    while(q--)
    {
        ri op=read();
        switch(op)
        {
        :{ri x=read(),a=read();modify_pre_sig(x,a);break;}
        :{ri x=read(),a=read();modify_pre(x,a);break;}
        :{ri x=read();printf("%lld\n",query_pre(x));break;}
        }
    }
    ;
}

只会做做小水题安慰下自己了QAQ