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Problem Code: ANUDTQ
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Given a directed tree with N nodes. You need to process M queries.
Each node has a key and a value, a node is referenced by its key. N nodes havekeys from 0 to N-1.
Root always has the key equal to 0. Queries can be of the following 4 types:
1. Given a key of a node present in the tree, append a child node to it. The new node gets the smallest never-used positive integer as its key. The value of the new node will be given in the input.
2. Given a key of a node(call it A) present in the tree. Add value to the value of all the nodes present in the subtree rooted at A.
3. Given a key of a node(call it A) present in the tree. Remove the subtree rooted at Afrom the tree.
4. Given a key of a node(call it A) present in the tree. Output the sum of values of all the nodes in subtree rooted at A.
Input is encoded:
Keys in M Queries of the input are encoded in the following way:
Let SPECIAL = 0 initially. Whenever a query of the type 4 occurs, SPECIAL is updated to the result of that query.
All the keys given in queries are encoded, in order to decode them you need to add SPECIAL to it.
That is, you are given encoded_key as input, to get key use the formula key =SPECIAL + encoded_key.
Input
First line of input has N, the number of nodes.
Second line has N integers, the values of given N nodes respectively.
Then, N-1 lines follow, each has two integers u, v. which specifies an edge from u to v.
Next line contains a single integer M, the number of queries.
Each query consists of 2 lines, First line has the the type of query, second line is as follows:
For queries 1 and 2, there are two integers, first one represents the key of a node in tree, second one represents the value.
For queries 3 and 4, there is a single integer, which represents the key of a node in tree.
Output
For each query of type 4, output the required answer.
Constraints
- 1 ≤ N ≤ 10^5
- 1 ≤ M ≤ 10^5
- 0 ≤ u < N
- 0 ≤ v < N, u is not equal to v
- All the keys in the input are valid
- If the type of the query is 3, the the key is nonzero.
- All the rest numbers in the input are in the range [-1000, 1000]
Example
Input: 2 10 20 0 1 4 4 0 1 -30 5 2 -30 1 4 -30
Output: 30 38
Explanation
Query #1
type = 4 encoded_key = 0
Initially SPECIAL = 0
key = encoded_key + SPECIAL = 0
value at 0 + value at 1 = 10 + 20 = 30 is the answer
Now SPECIAL is updated to 30
Query #2
type = 1 encoded_key = -30 value = 5
SPECIAL = 30
key = encoded_key + SPECIAL = -30 + 30 = 0
So we add a child node(with the key = 2) to the node with the key 0, the child node gets the value of 5
EDITS MADE:
Firstly, sorry for the mistake.
Old : Let SPECIAL = 0 initially. Whenever a query of the type 4 occurs, SPECIAL is increased by the result of that query.
New : Let SPECIAL = 0 initially. Whenever a query of the type 4 occurs, SPECIAL is updated to the result of that query.
题意：给出一颗由有向边构成的树，每个节点有编号、权值
操作1：给编号为A的点新增一个权值为val的叶子节点
操作2：给以编号A为根的子树权值全体+val
操作3：删除以编号为A为根的子树
操作4：询问以编号A为根的子树的权值和
碰到子树常用dfs序解决，但这题是动态增删，dfs序难以完成
所以 用splay维护括号序列 来解决子树问题
具体来说就是每加一个点，实际往splay中加一对括号
查找子树时，找到根节点的左右括号，就确定了子树范围
#include<cstdio> #define lc ch[x][0] #define rc ch[x][1] #define N 401001 using namespace std; int fa[N],ch[N][2],l[N],r[N],siz[N],a[N]; int front[N],to[N],next[N],tot,cnt; long long sum[N],tag[N],key[N],special; int n,m; int read() { int x=0;int f=1; char c=getchar(); while(c<'0'||c>'9') { if(c=='-') f=-1; c=getchar(); } while(c>='0'&&c<='9') { x=x*10+c-'0'; c=getchar(); } return x*f; } long long read2() { long long x=0;int f=1; char c=getchar(); while(c<'0'||c>'9') { if(c=='-') f=-1; c=getchar(); } while(c>='0'&&c<='9') { x=x*10+c-'0'; c=getchar(); } return x*f; } struct SPLAY { void up(int x) { siz[x]=siz[ch[x][0]]+siz[ch[x][1]]+1; sum[x]=sum[ch[x][0]]+sum[ch[x][1]]+key[x]; } void accumu(int x,long long val) { key[x]+=val; sum[x]+=val*siz[x]; tag[x]+=val; } void down(int x) { if(lc) accumu(lc,tag[x]); if(rc) accumu(rc,tag[x]); tag[x]=0; } bool getson(int x) { return ch[fa[x]][1]==x; } void rotate(int x) { int y=fa[x],z=fa[y],k=ch[y][1]==x; if(z) ch[z][ch[z][1]==y]=x; fa[x]=z; ch[y][k]=ch[x][k^1]; ch[x][k^1]=y; fa[y]=x; if(ch[y][k]) fa[ch[y][k]]=y; up(y); } void splay(int x,int g=0) { while(fa[x]!=g) { int y=fa[x],z=fa[y]; if(tag[z]) down(z); if(tag[y]) down(y); if(tag[x]) down(x); if(z!=g) rotate(getson(x)==getson(y) ? y : x); rotate(x); up(x); } } int find_pre(int y) { splay(y); int x=ch[y][0]; while(rc) x=rc; return x; } int find_suf(int y) { splay(y); int x=ch[y][1]; while(lc) x=lc; return x; } void insert(int x,int f,int val) { int pre=l[f],suf=find_suf(l[f]); splay(pre); splay(suf,pre); l[x]=++tot; r[x]=++tot; ch[suf][0]=l[x]; fa[l[x]]=suf; ch[l[x]][1]=r[x]; fa[r[x]]=l[x]; siz[l[x]]=siz[r[x]]=1; sum[l[x]]=key[l[x]]=val; sum[r[x]]=key[r[x]]=val; up(l[x]); up(suf); up(pre); } void add(int x,int val) { int pre=find_pre(l[x]); int suf=find_suf(r[x]); splay(pre); splay(suf,pre); accumu(ch[suf][0],val); up(suf); up(pre); } void del(int x) { int pre=find_pre(l[x]); int suf=find_suf(r[x]); splay(pre); splay(suf,pre); ch[suf][0]=0; up(suf); up(pre); } void query(int x) { int pre=find_pre(l[x]); int suf=find_suf(r[x]); splay(pre); splay(suf,pre); special=sum[ch[suf][0]]/2; printf("%lld ",special); } }Splay; struct TREE { void add(int u,int v) { to[++cnt]=v; next[cnt]=front[u]; front[u]=cnt; } void dfs(int u,int f) { Splay.insert(u,f,a[u]); for(int i=front[u];i;i=next[i]) dfs(to[i],u); } }Tree; int main() { n=read(); for(int i=1;i<=n;i++) a[i]=read(); int u,v; for(int i=1;i<n;i++) { u=read(); v=read(); Tree.add(++u,++v); } l[0]=++tot; r[0]=++tot; siz[1]=2; siz[2]=1; fa[2]=1; ch[1][1]=2; Tree.dfs(1,0); m=read(); int op; long long x; while(m--) { op=read(); x=read2(); x+=special; x++; if(op==1) { u=read(); Splay.insert(++n,x,u); } else if(op==2) { u=read(); Splay.add(x,u); } else if(op==3) Splay.del(x); else Splay.query(x); } }