BZOJ4765 普通计算姬（分块+树状数组）

　　对节点按编号分块。设f[i][j]为修改j号点对第i块的影响，计算f[i][]时dfs一遍即可。记录每一整块的sum。修改时对每一块直接更新sum，同时用dfs序上的树状数组维护子树和。查询时累加整块区间的sum，剩余部分bit上暴力查询。分析一下复杂度。设块大小为k，计算f数组的复杂度为O(n²/k)，修改复杂度为O(nm/k+mlogn)，查询复杂度O(nm/k+mklogn)。不妨设nm同阶，则k=sqrt(n/logn)时最优，总复杂度O(n·sqrt(nlogn))。然而真的这样的话f空间不够反正直接开sqrt(n)就过了。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll unsigned long long
#define N 100010
#define BLOCK 320
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,m,p[N],dfn[N],size[N],root,t,cnt;
int block,num,L[BLOCK],R[BLOCK],pos[N],f[BLOCK][N];
ll sum[BLOCK],tree[N],a[N];
struct data{int to,nxt;
}edge[N<<1];
void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}
void add(int k,int x){while (k<=n) tree[k]+=x,k+=k&-k;}
ll query(int k){ll s=0;while (k) s+=tree[k],k-=k&-k;return s;}
void calc(int k,int x,int from,int cnt)
{
    if (k>=L[x]&&k<=R[x]) cnt++;
    f[x][k]=cnt;
    for (int i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=from) calc(edge[i].to,x,k,cnt);
}
void dfs(int k,int from)
{
    dfn[k]=++cnt;size[k]=1;
    for (int i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=from) dfs(edge[i].to,k),size[k]+=size[edge[i].to];
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4765.in","r",stdin);
    freopen("bzoj4765.out","w",stdout);
    const char LL[]="%I64u
";
#else
    const char LL[]="%llu
";
#endif
    n=read(),m=read();
    for (int i=1;i<=n;i++) a[i]=read();
    block=sqrt(n);num=(n-1)/block+1;
    for (int i=1;i<=num;i++)
    {
        L[i]=(i-1)*block+1,R[i]=min(n,i*block);
        for (int j=L[i];j<=R[i];j++)
        pos[j]=i;
    }
    for (int i=1;i<=n;i++)
    {
        int x=read(),y=read();
        if (!x) root=y;
        else addedge(x,y),addedge(y,x);
    }
    for (int i=1;i<=num;i++) calc(root,i,root,0);
    dfs(root,root);
    for (int i=1;i<=n;i++) add(dfn[i],a[i]);
    for (int i=1;i<=n;i++) sum[pos[i]]+=query(dfn[i]+size[i]-1)-query(dfn[i]-1);
    while (m--)
    {
        int op=read();
        if (op==1)
        {
            int x=read(),y=read();
            y-=a[x];
            add(dfn[x],y);
            for (int i=1;i<=num;i++) sum[i]+=(ll)f[i][x]*y;
            a[x]+=y;
        }
        else
        {
            int l=read(),r=read();
            ll ans=0;
            if (pos[l]==pos[r])
            {
                for (int i=l;i<=r;i++)
                ans+=query(dfn[i]+size[i]-1)-query(dfn[i]-1);
            }
            else
            {
                for (int i=pos[l]+1;i<pos[r];i++)
                ans+=sum[i];
                for (int i=l;i<=R[pos[l]];i++)
                ans+=query(dfn[i]+size[i]-1)-query(dfn[i]-1);
                for (int i=L[pos[r]];i<=r;i++)
                ans+=query(dfn[i]+size[i]-1)-query(dfn[i]-1);
            }
            printf(LL,ans);
        }
    }
    return 0;
}