HDU4918 Query on the subtree 点分治+树状数组
bobo has a tree, whose vertices are conveniently labeled by 1,2,…,n. At the very begining, the i-th vertex is assigned with weight w i.
There are q operations. Each operations are of the following 2 types:
Change the weight of vertex v into x (denoted as "! v x"),
Ask the total weight of vertices whose distance are no more than d away from vertex v (denoted as "? v d").
Note that the distance between vertex u and v is the number of edges on the shortest path between them.
There are q operations. Each operations are of the following 2 types:
Change the weight of vertex v into x (denoted as "! v x"),
Ask the total weight of vertices whose distance are no more than d away from vertex v (denoted as "? v d").
Note that the distance between vertex u and v is the number of edges on the shortest path between them.
InputThe input consists of several tests. For each tests:
The first line contains n,q (1≤n,q≤10 5). The second line contains n integers w 1,w 2,…,w n (0≤w i≤10 4). Each of the following (n - 1) lines contain 2 integers a i,b i denoting an edge between vertices a i and b i (1≤a i,b i≤n). Each of the following q lines contain the operations (1≤v≤n,0≤x≤10 4,0≤d≤n).
OutputFor each tests:
For each queries, a single number denotes the total weight.Sample Input
4 3 1 1 1 1 1 2 2 3 3 4 ? 2 1 ! 1 0 ? 2 1 3 3 1 2 3 1 2 1 3 ? 1 0 ? 1 1 ? 1 2
Sample Output
3 2 1 6 6
题意:给你一棵树,N个点,每个点一个权值,然后Q组操作(共两种),第一种是求导一个节点距离不超过d的所有点的权值和是多少;第二种操作时,修改一个点的权值;
题解:多次进行操作1。此时不能每次都O(NlogN)了,太慢了。我们考虑到对于点分治,树的重心一共有logN层,第一层为整棵树的重心,第二层为第一层重心的子树的重心,以此类推,每次至少分成两个大小差不多的子树,所以
一共有logN层。而且,对于一个点,他最多只属于logN个子树,也就是最多只属于logN个重心。所以我们可以预处理出每个点所属于的重心以及到这些重心的距离,以每个重心建树状数组,每个点按照到重心的距离插入到树状数组中,
然后每次查询到u距离不超过d的点的个数就通过树状数组求前缀和得到。假设一个重心x到u的距离为dis,那么便统计到重心x距离不超过d-dis的点的个数,这个过程我们称之为“借力”,本身能力有限,所以需要借助x的影响力。因为
如果这个重心被u借力了,那么这个重心的子重心一定也被借力,由于相邻被借力的两个重心x、y所统计的点会有重复,所以我们需要去重。去重的话我们就通过对每个节点再开一个v对x的树状数组,这个树状数组的意义为:重心x的子
树v的重心为y时,子树v中每个点到x的距离为下标建立的树状数组。因为重心x与重心y交集的部分,重心x包括的部分重心y一定包括,所以统计的时候减去v对x的树状数组中距x不超过d-dis的点的个数即可。访问u所属与的所有重心,
挨个借力,同时去重,便能得到距离u不超过d的点的个数。因为重心最多logN层,每个树状数组最多N个点,logN复杂度的统计,所以每次查询复杂度O(logN*logN)。我们最多为每个节点开2个树状数组,而且每一层所有树状数组的
大小相加不超过N,所以树状数组的占用空间为O(2NlogN)。
在上面的基础上稍做扩充。预处理的时候插入树状数组的就是该点的权值,查询依旧是统计前缀和。修改点权值的时候,便是和查询一样,在u距重心x距离d的位置在x的树状数组中修改u的权值,同时修改u属于重心x的子树v的v对x的树
状数组中相同位置的值。复杂度和查询一样为O(logN*logN)。
参考代码:
#include<bits/stdc++.h> using namespace std; #define pii pair<int,int> #define mkp make_pair #define lowbit(x) (x&-x) typedef long long ll; const int INF=0x3f3f3f3f; const int maxn=1e5+10; int n,q,w[maxn]; char op[2]; struct MSG{ int id1,id2; int dep; } msg[maxn][17]; struct Edge{ int v,nxt; } edge[maxn<<1]; int vis[maxn],head[maxn],tot; int root,siz[maxn],mx[maxn],fa[maxn],S; int maxfloor[maxn],idn,L[maxn<<1],R[maxn<<1]; int c[maxn*17]; void Init() { S=n;idn=0; tot=root=0; memset(c,0,sizeof c); memset(vis,0,sizeof vis); memset(w,0,sizeof w); memset(head,-1,sizeof head); memset(msg,0,sizeof msg); } void AddEdge(int x,int y) { edge[tot].v=y; edge[tot].nxt=head[x]; head[x]=tot++; } void Add(int l,int r,int pos,int val) { while(pos<r-l) c[l+pos]+=val,pos+=lowbit(pos); } int Sum(int l,int r,int len) { if(len<1) return 0; if(len>r-l-1) len=r-l-1; int res=0; while(len) res+=c[l+len],len-=lowbit(len); return res; } void getroot(int u,int fa) { siz[u]=1;mx[u]=0; for(int i=head[u];~i;i=edge[i].nxt) { int v=edge[i].v; if(vis[v]||v==fa) continue; getroot(v,u); siz[u]+=siz[v]; mx[u]=max(mx[u],siz[v]); } mx[u]=max(mx[u],S-siz[u]); if(mx[u]<mx[root]) root=u; } int getmaxdep(int u,int fa) { int res=1; for(int i=head[u];~i;i=edge[i].nxt) { int v=edge[i].v; if(vis[v]||v==fa) continue; res=max(res,1+getmaxdep(v,u)); } return res; } void dfs(int u,int fa,int deep,int id,int flor,int tp) { if(!tp) msg[u][flor].id1=id; else msg[u][flor].id2=id; msg[u][flor].dep=deep; Add(L[idn],R[idn],deep,w[u]); for(int i=head[u];~i;i=edge[i].nxt) { int v=edge[i].v; if(!vis[v]&&v!=fa) dfs(v,u,deep+1,id,flor,tp); } } void solve(int u,int s,int flor) { vis[u]=1; maxfloor[u]=flor; idn++; L[idn]=R[idn-1]; R[idn]=L[idn]+getmaxdep(u,0)+1; dfs(u,0,1,idn,flor,0); msg[u][flor].id2=-1; for(int i=head[u];~i;i=edge[i].nxt) { int v=edge[i].v; if(vis[v]) continue; idn++;L[idn]=R[idn-1]; R[idn]=L[idn]+getmaxdep(v,u)+2; dfs(v,u,2,idn,flor,1); } for(int i=head[u];~i;i=edge[i].nxt) { int v=edge[i].v; if(vis[v]) continue; S=siz[v]; root=0; if(siz[v]>siz[u]) S=s-siz[u]; getroot(v,u); solve(root,siz[v],flor+1); } } int main() { while(~scanf("%d%d",&n,&q)) { Init(); for(int i=1;i<=n;++i) scanf("%d",w+i); for(int i=1;i<n;++i) { int x,y; scanf("%d%d",&x,&y); AddEdge(x,y);AddEdge(y,x); } mx[root]=INF; getroot(1,0); solve(1,S,0); while(q--) { int x,y,ans; scanf("%s%d%d",&op,&x,&y); if(op[0]=='?') { ans=0; for(int f=0;f<=maxfloor[x];++f) { int id1=msg[x][f].id1; int id2=msg[x][f].id2; int dep=msg[x][f].dep; ans+=Sum(L[id1],R[id1],y+2-dep); if(id2!=-1) ans-=Sum(L[id2],R[id2],y+2-dep); } printf("%d\n",ans); } else { for(int f=0;f<=maxfloor[x];++f) { int id1=msg[x][f].id1; int id2=msg[x][f].id2; int dep=msg[x][f].dep; Add(L[id1],R[id1],dep,y-w[x]); if(id2!=-1) Add(L[id2],R[id2],dep,y-w[x]); } w[x]=y; } } } return 0; }