BZOJ4999 This Problem Is Too Simple!（树上差分+dfs序+树状数组）

　　对每个权值分别考虑。则只有单点加路径求和的操作。树上差分转化为求到根的路径和，子树加即可。再差分后bit即可。注意树上差分中根的父亲是0，已经忘了是第几次因为这个挂了。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int n,m,a[N],p[N],dfn[N],size[N],deep[N],fa[N][],tree[N],ans[N*],t,cnt;
struct data{int to,nxt;
}edge[N<<];
struct data2
{
    int op,i,j,x,id;
    bool operator <(const data2&a) const
    {
        return x<a.x||x==a.x&&id<a.id;
    }
}q[N*];
void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}
void dfs(int k)
{
    dfn[k]=++cnt,size[k]=;
    for (int i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=fa[k][])
    {
        deep[edge[i].to]=deep[k]+;
        fa[edge[i].to][]=k;
        dfs(edge[i].to);
        size[k]+=size[edge[i].to];
    }
}
void add(int k,int x){while (k<=n) tree[k]+=x,k+=k&-k;}
int query(int k) {int s=;while (k) s+=tree[k],k-=k&-k;return s;}
int lca(int x,int y)
{
    if (deep[x]<deep[y]) swap(x,y);
    for (int j=;~j;j--) if (deep[fa[x][j]]>=deep[y]) x=fa[x][j];
    if (x==y) return x;
    for (int j=;~j;j--) if (fa[x][j]!=fa[y][j]) x=fa[x][j],y=fa[y][j];
    return fa[x][];
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4999.in","r",stdin);
    freopen("bzoj4999.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read();
    for (int i=;i<=n;i++) a[i]=read();
    for (int i=;i<n;i++)
    {
        int x=read(),y=read();
        addedge(x,y),addedge(y,x);
    }
    fa[][]=;dfs();
    for (int j=;j<;j++)
        for (int i=;i<=n;i++)
        fa[i][j]=fa[fa[i][j-]][j-];
    int x=;
    for (int i=;i<=n;i++) x++,q[x].op=,q[x].i=i,q[x].j=,q[x].x=a[i],q[x].id=x;
    for (int i=;i<=m;i++)
    {
        char c=getc();
        if (c=='C')
        {
            x++,q[x].op=,q[x].i=read(),q[x].j=,q[x].x=read(),q[x].id=x;
            x++,q[x].op=,q[x].i=q[x-].i,q[x].j=-,q[x].x=a[q[x].i],q[x].id=x;
            a[q[x].i]=q[x-].x;
        }
        else x++,q[x].op=,q[x].i=read(),q[x].j=read(),q[x].x=read(),q[x].id=x;
    }
    m=x;
    sort(q+,q+m+);memset(ans,,sizeof(ans));
    for (int i=;i<=m;i++)
    {
        int t=i;
        while (t<m&&q[t+].x==q[i].x) t++;
        for (int j=i;j<=t;j++)
        if (q[j].op)
        {
            int l=lca(q[j].i,q[j].j);
            ans[q[j].id]=query(dfn[q[j].i])+query(dfn[q[j].j])-query(dfn[l])-query(dfn[l==?:fa[l][]]);
        }
        else add(dfn[q[j].i],q[j].j),add(dfn[q[j].i]+size[q[j].i],-q[j].j);
        for (int j=i;j<=t;j++)
        if (q[j].op==) add(dfn[q[j].i],-q[j].j),add(dfn[q[j].i]+size[q[j].i],q[j].j);
        i=t;
    }
    for (int i=;i<=m;i++) if (ans[i]>=) printf("%d\n",ans[i]);
    return ;
}