2018沈阳网络预选赛 kachang

Ka Chang

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1000ms
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Given a rooted tree ( the root is node 11) of NN nodes. Initially, each node has zero point.

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

1\ L\ X1 L X: Increase points by XX of all nodes whose depth equals LL ( the depth of the root is zero ). (x \leq 10^8)(x≤108)

2\ X2 X: Output sum of all points in the subtree whose root is XX.

Input

Just one case.

The first lines contain two integer, N,QN,Q. (N \leq 10^5, Q \leq 10^5)(N≤105,Q≤105).

The next n-1n−1 lines: Each line has two integer aa,bb, means that node aa is the father of node bb. It's guaranteed that the input data forms a rooted tree and node 11 is the root of it.

The next QQ lines are queries.

Output

For each query 22, you should output a number means answer.

样例输入复制

样例输出复制

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题目来源

ACM-ICPC 2018 沈阳赛区网络预赛

题目大意：给出一个有根树，有两种操作：

1.第x层的所有点权+y

2.查询以x为根的子树的权值之和。

解题思路：可以想到按照dfs序二分每一层，但是这样复杂度很高。

正解：对于某一层，如果当前曾的点数比较少，则直接在树状数组里边更改，由于查询子树的时候是连续的，所以我们可以直接得到以某个点为根节点的点数较小的层的权值之和。对于点数较多的层，我们就直接在这些层中二分。

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define sca(x) scanf("%d",&x)
#define N 200005
#define pb(x) push_back(x)
#define lowb(x) (x&(-x))struct node
{int to,nt;
}g[N*2];
int tot;
int head[N],in[N],ou[N],sz[N];
LL tag[N];
vector<int>big;void addedg(int x,int y)
{g[tot].to=y;g[tot].nt=head[x];head[x]=tot++;
}vector<int>V[N];int vis[N];
int id;
void dfs(int u,int dep)
{vis[u]=1;in[u]=++id;V[dep].pb(id);for(int i=head[u];i+1;i=g[i].nt){int v=g[i].to;if(vis[v])continue;dfs(v,dep+1);}ou[u]=id;
}struct BIT
{LL c[N];int maxn;void init(int n){maxn=n;}LL ask(int x){LL ans=0;for(int i=x;i;i-=lowb(i))ans+=c[i];return ans;}void upd(int x,LL val){for(int i=x;i<=maxn;i+=lowb(i))c[i]+=val;}}bit;int main()
{int n,q;sca(n),sca(q);memset(head,-1,sizeof(head));for(int i=1;i<n;i++){int x,y;sca(x),sca(y);addedg(x,y);addedg(y,x);}dfs(1,0);int up=sqrt(n);bit.init(n);for(int i=0;i<n;i++){sz[i]=V[i].size();if(sz[i]>up)big.pb(i);}for(int i=0;i<q;i++){int op;sca(op);if(op==1){int l;LL x;scanf("%d%lld",&l,&x);if(sz[l]<=up){for(int j=0;j<sz[l];j++){bit.upd(V[l][j],x);}}else tag[l]+=x;}else{int x;scanf("%d",&x);LL ans1=bit.ask(ou[x])-bit.ask(in[x]-1);for(int j=0;j<big.size();j++){int _id=big[j];ans1+=1LL*(upper_bound(V[_id].begin(),V[_id].end(),ou[x])-lower_bound(V[_id].begin(),V[_id].end(),in[x]))*tag[_id];}printf("%lld\n",ans1);}}
}