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) const int N=5000000+10; 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=0;rb y=1; while(ch!='-' && (ch<'0' || ch>'9'))ch=gc; if(ch=='-')ch=gc,y=0; while(ch>='0' && ch<='9')x=(x<<1)+(x<<3)+(ch^'0'),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;} void dfs1(ri x,ri fat){sz[x]=1;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)>>1; tr[ls(x)].tag+=tr[x].tag;tr[ls(x)].dat+=tr[x].tag*(mid-l+1); tr[rs(x)].tag+=tr[x].tag;tr[rs(x)].dat+=tr[x].tag*(r-mid); tr[x].tag=0; } } 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)>>1; build(ls(x),l,mid);build(rs(x),mid+1,r); pushup(x); } void modify(ri x,ri l,ri r,ri to_l,ri to_r,ri dat) { if(to_l<=l && r<=to_r){tr[x].tag+=dat;tr[x].dat+=dat*(r-l+1);return;} pushdown(x,l,r); ri mid=(l+r)>>1; if(mid>=to_l)modify(ls(x),l,mid,to_l,to_r,dat); if(mid<to_r)modify(rs(x),mid+1,r,to_l,to_r,dat); pushup(x); } int query(ri x,ri l,ri r,ri to_l,ri to_r) { if(l>r)return 0; if(to_l<=l && r<=to_r)return tr[x].dat; pushdown(x,l,r); ri mid=(l+r)>>1,ret=0; if(mid>=to_l)ret+=query(ls(x),l,mid,to_l,to_r); if(mid<to_r)ret+=query(rs(x),mid+1,r,to_l,to_r); return ret; } il void modify_pre_sig(ri x,ri dat){modify(1,1,n,dfn[x],dfn[x],dat);} il void modify_pre(ri x,ri dat){modify(1,1,n,dfn[x],low[x],dat);} il int query_pre(ri x){ri ret=0;while(x)ret+=query(1,1,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,1,n)val[i]=read();rp(i,1,n-1){ri x=read(),y=read();ad(x,y);}dfs1(1,0);dfs2(1,1);build(1,1,n); while(q--) { ri op=read(); switch(op) { case 1:{ri x=read(),a=read();modify_pre_sig(x,a);break;} case 2:{ri x=read(),a=read();modify_pre(x,a);break;} case 3:{ri x=read();printf("%lld\n",query_pre(x));break;} } } return 0; }