Our protagonist is the handsome human prince Aragorn comes from The Lord of the Rings. One day Aragorn finds a lot of enemies who want to invade his kingdom. As Aragorn knows, the enemy has N camps out of his kingdom and M edges connect them. It is guaranteed that for any two camps, there is one and only one path connect them. At first Aragorn know the number of enemies in every camp. But the enemy is cunning , they will increase or decrease the number of soldiers in camps. Every time the enemy change the number of soldiers, they will set two camps C1 and C2. Then, for C1, C2 and all camps on the path from C1 to C2, they will increase or decrease K soldiers to these camps. Now Aragorn wants to know the number of soldiers in some particular camps real-time.

Multiple test cases, process to the end of input.

For each case, The first line contains three integers N, M, P which means there will be N(1 ≤ N ≤ 50000) camps, M(M = N-1) edges and P(1 ≤ P ≤ 100000) operations. The number of camps starts from 1.

The next line contains N integers A1, A2, ...AN(0 ≤ Ai ≤ 1000), means at first in camp-i has Ai enemies.

The next M lines contains two integers u and v for each, denotes that there is an edge connects camp-u and camp-v.

The next P lines will start with a capital letter ‘I‘, ‘D‘ or ‘Q‘ for each line.

‘I‘, followed by three integers C1, C2 and K( 0≤K≤1000), which means for camp C1, C2 and all camps on the path from C1 to C2, increase K soldiers to these camps.

‘D‘, followed by three integers C1, C2 and K( 0≤K≤1000), which means for camp C1, C2 and all camps on the path from C1 to C2, decrease K soldiers to these camps.

‘Q‘, followed by one integer C, which is a query and means Aragorn wants to know the number of enemies in camp C at that time.

For each query, you need to output the actually number of enemies in the specified camp.

3 2 5
1 2 3
2 1
2 3
I 1 3 5
Q 2
D 1 2 2
Q 1 
Q 3

7
4
8
Hint
1.The number of enemies may be negative.

2.Huge input, be careful. 
 

一开始提交，爆栈，后在网上看到：

题意：给一棵树，并给定各个点权的值，然后有3种操作：

I C1 C2 K: 把C1与C2的路径上的所有点权值加上K

D C1 C2 K：把C1与C2的路径上的所有点权值减去K

Q C：查询节点编号为C的权值

 

分析：典型的树链剖分题目，先进行剖分，然后用线段树去维护即可，注意HDU的OJ采用Windows系统，容易爆栈，所以在代码

前面加上：#pragma comment(linker, "/STACK:1024000000,1024000000")进行手动扩栈。

#pragma comment(linker, "/STACK:1024000000,1024000000") //因OJ采用Windows系统，要加入这一行用于 进行手动扩栈，这样就不会引起爆栈
#include<stdio.h>
#include<string.h>
#include<vector>
using namespace std;
const int N = 50015;

vector<int>mapt[N];
int fath[N],deep[N],top[N],num[N],son[N],p[N],pos;

void init(int n)
{
    pos=0;
    //memset(son,-1,sizeof(son));
    for(int i=0;i<=n;i++)
        mapt[i].clear();
}
void dfs(int u,int pre,int d)
{
    fath[u]=pre; num[u]=1; deep[u]=d; son[u]=-1;
    int k=mapt[u].size();
    for(int i=k-1;i>=0;--i)
    {
        int v=mapt[u][i];
        if(v==pre)continue;
        dfs(v,u,d+1);
        num[u]+=num[v];
        if(son[u]==-1||num[son[u]]<num[v])
            son[u]=v;
    }
}
void getpos(int u,int root)
{
    top[u]=root;
    p[u]=++pos;//从1开始
    if(son[u]==-1)
        return ;
    getpos(son[u],root);
    int k=mapt[u].size();
    for(int i=k-1;i>=0;--i)
    {
        int v=mapt[u][i];
        if(son[u]==v||v==fath[u])
            continue;
        getpos(v,v);
    }
}

struct tree
{
    int allnum,addnum;
}root[N*3];
int a[N];

void build(int l,int r,int k)
{
    root[k].addnum=0;
    if(l==r){
        root[k].allnum=a[l]; return ;
    }
    int mid=(l+r)/2;
    build(l,mid,k<<1);
    build(mid+1,r,(k<<1)|1);
    root[k].allnum=root[k<<1].allnum+root[(k<<1)|1].allnum;
}
void upson(int l,int r,int k)
{
    int mid=(l+r)/2;
    if(root[k].addnum){
        root[k<<1].allnum+=(mid-l+1)*root[k].addnum;
        root[k<<1].addnum+=root[k].addnum;

        root[(k<<1)|1].allnum+=(r-mid)*root[k].addnum;
        root[(k<<1)|1].addnum+=root[k].addnum;
        root[k].addnum=0;
    }
}
void update(int l,int r,int k,int L,int R,int c)
{
    if(L<=l&&r<=R)
    {
        root[k].allnum+=(r-l+1)*c; root[k].addnum+=c; return ;
    }
    int mid=(l+r)/2;
    upson(l,r,k);
    if(L<=mid) update(l,mid,k<<1,L,R,c);
    if(mid<R)  update(mid+1,r,(k<<1)|1,L,R,c);
    root[k].allnum=root[k<<1].allnum+root[(k<<1)|1].allnum;
}
int query(int l,int r,int k,int id)
{
    if(l==r)
        return root[k].allnum;
    upson(l,r,k);
    int mid=(l+r)/2;
    if(id<=mid)
        return query(l,mid,k<<1,id);
    else return query(mid+1,r,(k<<1)|1,id);
}

void swp(int &a,int &b)
{
    int tt=a; a=b; b=tt;
}
void Operat(int u,int v,int c)
{
    int fu=top[u], fv=top[v];
    while(fu!=fv)
    {
        if(deep[fu]<deep[fv])
        {
            swp(fu,fv); swp(u,v);
        }
        update(1,pos,1,p[fu],p[u],c);
        u=fath[fu]; fu=top[u];
    }
    if(deep[u]>deep[v])
        swp(u,v);
    update(1,pos,1,p[u],p[v],c);//点权更新与边权更新的区别之一:点p[u]，边p[son[u]]
}

int main()
{
    int n,m,q,val[N];
    while(scanf("%d%d%d",&n,&m,&q)>0)
    {
        init(n);
        for(int i=1;i<=n;i++)
            scanf("%d",&val[i]);
        while(m--)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            mapt[u].push_back(v);
            mapt[v].push_back(u);
        }
        dfs(1,1,1);
        getpos(1,1);
        for(int i=1;i<=n;i++)
            a[p[i]]=val[i]; //一定要注意转换成在线段树上的对应位置
        pos=n;
        build(1,pos,1);

        while(q--)
        {
            int u,v,c;
            char ch[10];
            scanf("%s%d",ch,&u);
            if(ch[0]=='Q')
                printf("%d\n",query(1,pos,1,p[u]));
            else
            {
                scanf("%d%d",&v,&c);
                if(ch[0]=='D') c=-c;
                Operat(u,v,c);
            }
        }
    }
    return 0;
}