标签：++ strong ase clipboard target sqrt data font res
Given a rooted tree ( the root is node 11 ) of NN nodes. Initially, each node has zero point.
Then, you need to handle QQ operations. 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.
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.
For each query 22, you should output a number means answer.
3 3
1 2
2 3
1 1 1
2 1
2 3
1
0
SOLUTION:
第一次写分块这种的暴力的方法qwq
发现本题 查询子树和 与 修改相同深度的点的val时相互矛盾的
那就考虑怎么暴力吧
即，如果修改的那个层数的节点小于某个T，那么就直接用树状数组+DFS序维护子树和，暴力单点更新。
如果 大于> ，那么就用懒惰标记的思想，先把加了多少 记录下来，等查询的时候再暴力查询该子树，深度为某个值的节点个数，然后乘以该懒惰标记即可。那么这个T多大呢？明显是sqrt(N)。
那么查询某子树，深度为某个值的节点个数怎么做呢？先把节点按照dfs序编号，然后记录某层有哪些节点，最后二分查询即可。
CODE:
#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+7;
typedef long long ll;
vector<int>G[maxn],dep[maxn],large;
int st[maxn],ed[maxn],tim,n;
ll tree[maxn],num[maxn];
void add(int pos,ll v)
{
for(;pos<=n;pos+=pos&-pos)
tree[pos]+=v;
}
ll query(int pos)
{
ll ans=0;
for(;pos;pos-=pos&-pos)
ans+=tree[pos];
return ans;
}
void dfs(int u,int fa,int depth) //时间戳
{
st[u]=++tim;
dep[depth].push_back(tim);
int sz=G[u].size();
for(int i=0;i<sz;i++)
if(G[u][i]!=fa)
dfs(G[u][i],u,depth+1);
ed[u]=tim;
}
int main()
{
int q;
scanf("%d%d",&n,&q);
for(int i=0;i<n-1;i++)
{
int u,v;
scanf("%d%d",&u,&v);
G[u].push_back(v);
G[v].push_back(u);
}
dfs(1,0,0);
int block=ceil(sqrt(n));
for(int i=0;i<n;i++)
if(dep[i].size()>block)
large.push_back(i);
while(q--)
{
int nood;
scanf("%d",&nood);
if(nood==1)
{
int depth;
ll v;
scanf("%d%lld",&depth,&v);
if(dep[depth].size()>block) num[depth]+=v;
else
{
for(int i=0;i<dep[depth].size();i++)
add(dep[depth][i],v);
}
}
else
{
int x;
scanf("%d",&x);
ll ans=query(ed[x])-query(st[x]-1);
for(int i=0;i<large.size();i++)
ans+=(upper_bound(dep[large[i]].begin(),dep[large[i]].end(),ed[x])-lower_bound(dep[large[i]].begin(),dep[large[i]].end(),st[x]))*num[large[i]];
printf("%lld\n",ans);
}
}
}
[ACM-ICPC 2018 沈阳网络赛] Ka Chang （dfs序+树状数组+分块）
标签：++ strong ase clipboard target sqrt data font res
原文地址：https://www.cnblogs.com/zhangbuang/p/11180202.html
