ACM-ICPC 2018 沈阳赛区网络预赛 J. Ka Chang(树状数组+分块)

Given a rooted tree ( the root is node 1 ) of N nodes. Initially, each node has zero point.

Then, you need to handle Q operations. There're two types:

1 L X: Increase points by X of all nodes whose depth equals L ( the depth of the root is zero ). (x≤10⁸)

2 X: Output sum of all points in the subtree whose root is X.

Input
Just one case.

The first lines contain two integer, N,Q(N≤10⁵,Q≤10⁵).

The next n−1 lines: Each line has two integer aaa,bbb, means that node aaa is the father of node b. It's guaranteed that the input data forms a rooted tree and node 1 is the root of it.

The next Q lines are queries.

Output
For each query 2, you should output a number means answer.

样例输入
3 3
1 2
2 3
1 1 1
2 1
2 3

样例输出
1
0

题意

给你一颗以1为根节点的数，有两个操作

1.层数为L的节点增加X

2.查询以X为根节点的子树总权和

题解

操作2很容易想到dfs维护树状数组

对于操作1，可以把层数按sqrt(n)分块，对于层数点数<=block的直接暴力更新，对于>block先保存进数组ans，查询的时候二分左右区间可以知道根的节点数，再乘ans，这样可以把单次操作的复杂度固定在sqrt(n)范围内

代码

 #include<bits/stdc++.h>
 using namespace std;

 #define LL long long

 const int maxn=1e5+;
 int s[maxn],e[maxn],tot,n;
 LL sum[maxn],ans[maxn];
 vector<int>G[maxn],deep[maxn],q;
 void dfs(int u,int fa,int d)
 {
     s[u]=++tot;
     deep[d].push_back(s[u]);
     for(auto v:G[u])
     {
         if(v==fa)continue;
         dfs(v,u,d+);
     }
     e[u]=tot;
 }
 int lowbit(int x)
 {
     return x&(-x);
 }
 void update(int x,int add)
 {
     for(int i=x;i<=n;i+=lowbit(i))
         sum[i]+=add;
 }
 LL query(int x)
 {
     LL ans=;
     for(int i=x;i>;i-=lowbit(i))
         ans+=sum[i];
     return ans;
 }
 int main()
 {
     int Q,op,L,x,u,v;
     scanf("%d%d",&n,&Q);
     int block=sqrt(n);
     for(int i=;i<n;i++)
     {
         scanf("%d%d",&u,&v);
         G[u].push_back(v);
         G[v].push_back(u);
     }
     dfs(,-,);
     for(int i=;i<n;i++)
         if(deep[i].size()>block)q.push_back(i);
     for(int i=;i<Q;i++)
     {
         scanf("%d",&op);
         if(op==)
         {
             scanf("%d%d",&L,&x);
             if(deep[L].size()>block)ans[L]+=x;
             else
             {
                 for(auto X:deep[L])
                     update(X,x);
             }
         }
         else
         {
             scanf("%d",&x);
             LL res=query(e[x])-query(s[x]-);
             for(auto pos:q)
                 res+=ans[pos]*(upper_bound(deep[pos].begin(),deep[pos].end(),e[x])-lower_bound(deep[pos].begin(),deep[pos].end(),s[x]));
             printf("%lld\n",res);
         }
     }
     return ;
 }