ACM-ICPC 2018 沈阳赛区网络预赛 J. Ka Chang (树分块)

题意:一个树,支持两种操作:1.将深度为L的节点权置加上X;2.求以x为根节点的子树上节点权置之和.根节点深度为0.
分析:考虑用树状数组维护节点权置,按dfs序下标查询.记录每个深度节点的个数.如果每次都暴力维护子树上每一层的节点,则会超时.
要用分块来解决.对于节点数量小于\(\sqrt{N}\)的层数,用树状数组维护;否则将该层记录,修改时单独记录.
查询时,答案分成两份:树状数组中维护的子树区间的和;以及属于根为所查节点x子树,且层号是超过\(\sqrt{N}\)的节点的权置和.后者查询时,对每一个超过\(\sqrt{N}\)的层号,求其层中节点在x子树区间中的节点权置之和.

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 2e5+5;
struct Edge{
    int v,next;
}edges[MAXN<<1];
int head[MAXN], tot,N,deep,dfs_cnt;
int L[MAXN],R[MAXN];
LL bit[MAXN<<1];
vector<int> dp[MAXN];
vector<int> bg;

void init(){
    memset(bit,0,sizeof(bit));
    memset(head,-1,sizeof(head));
    tot =0;
    deep = dfs_cnt = 0;
}

void AddEdge(int u,int v){
    edges[tot] = (Edge){v,head[u]};
    head[u] = tot++;
}

void add(int pos,LL val){
    for(int i=pos; i<=N; i+= (i&-i)) bit[i] += val;
}

LL sum(int pos)
{
    LL res=0;
    for(int i=pos; i ;i-= (i&-i)) res+= bit[i];
    return res;
}

void dfs(int u,int fa,int d)
{
    L[u] = ++ dfs_cnt;
    deep = max(deep,d);
    dp[d].push_back(L[u]);
    for(int i=head[u];~i;i=edges[i].next){
        int v = edges[i].v;
        if(v!=fa) dfs(v,u,d+1);
    }
    R[u] = dfs_cnt;
}
LL ans[MAXN];

int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    int Q,op,u,v,tmp;
    scanf("%d %d",&N, &Q);
    init();
    memset(ans,0,sizeof(ans));
    bg.clear();
    for(int i=0;i<=N;++i) dp[i].clear();

    for(int i=1;i<=N-1;++i){
        scanf("%d %d",&u,&v);
        AddEdge(u,v), AddEdge(v,u);
    }
    dfs(1,-1,0);
    int block = sqrt(1.0*N);
    for(int i=0;i<=deep;++i){           //该层节点大于sqrt(n)的单独保存
        if(dp[i].size()>block) bg.push_back(i);
    }
    while(Q--){
        scanf("%d",&op);
        if(op==1){
            int lev; LL x;
            scanf("%d %lld",&lev, &x);
            if(dp[lev].size()>block){       //节点数大的打上标记
                ans[lev] += x;
            }
            else{                           //该层节点数小的直接暴力更新.
                int siz = dp[lev].size();
                for(int i=0;i<siz;++i){
                    add(dp[lev][i],x);
                }
            }
        }
        else{
            scanf("%d",&u);
            LL res = sum(R[u]) - sum(L[u]-1);   //先求层数小的节点和
            int siz = bg.size();
            for(int i=0;i<siz;++i){
                int lev = bg[i];
                res += (upper_bound(dp[lev].begin(),dp[lev].end(),R[u])-
                lower_bound(dp[lev].begin(),dp[lev].end(),L[u]))*ans[lev];
            }
            printf("%lld\n",res);
        }
    }
    return 0;
}