题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=3966

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Aragorn’s Story

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10824 Accepted Submission(s): 2838

Problem Description
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.

Input
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,increase K soldiers to these camps.

‘D’,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.

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

Sample Input
3 2 5
1 2 3
2 1
2 3
I 1 3 5
Q 2
D 1 2 2
Q 1
Q 3

Sample Output
7
4
8
Hint

1.The number of enemies may be negative.

2.Huge input,be careful.

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题目大意：
给一棵树 ，n个节点，m条变，q个操作，
操作有三种，
1） I C1 C2 K 将点c1 到点c2的每个点 权值+k
2） D C1 C2 K 将点c1 到点c2的每个点 权值-k
3） Q C 查询当前点C的权值

解题思路：
还是树链剖分的入门 ，

但是这题简单在 只要用树状数组维护就行了，

坑比的我居然折在了输入上，

也后输入字符一论用$字符串%s$输入 ，

附本题代码
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#include<bits/stdc++.h>
using namespace std;

typedef long long int LL;

const int    INF = (~(@H_403_139@1<<@H_403_139@31));
const int    N   = @H_403_139@50000+@H_403_139@7;
const double eps = @H_403_139@1e-7;

inline int read(){
    int x=@H_403_139@0,f=@H_403_139@1;char ch = getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-@H_403_139@1;ch=getchar();}
    while('0'<=ch&&ch<='9'){x=(x<<@H_403_139@3)+(x<<@H_403_139@1)+ch-'0';ch=getchar();}
    return x*f;
}
/****************************************************/

int w[N],n,m,q;
vector<int >G[N];

void add(int u,int v){
    G[u].push_back(v);
    G[v].push_back(u);
}

int fa[N],dep[N],son[N],sz[N];
void dfs1(int u,int f,int d){
    fa[u]=f,dep[u]=d,sz[u]=@H_403_139@1;
    int gz=G[u].size();
    for(int to,i=@H_403_139@0;i<gz;i++){
        to=G[u][i];
        if(to==f) continue;
        dfs1(to,u,d+@H_403_139@1);
        sz[u]+=sz[to];
        if(sz[son[u]]<sz[to])son[u]=to;
    }
}

int top[N],tree[N],pre[N],cnt;
void dfs2(int u,int tp){
    top[u]=tp,tree[u]=++cnt,pre[tree[u]]=u;
    if(son[u]) dfs2(son[u],tp);
    int gz=G[u].size();
    for(int to,i=@H_403_139@0;i<gz;i++){
        to=G[u][i];
        if(to==son[u]||to==fa[u])continue;
        dfs2(to,to);
    }
}

int sum[N];
#define lowbit(x) (x&-x)
void update(int index,int val){
    for(int i=index;i<=n;i+=lowbit(i))sum[i]+=val;
}
int getSum(int index){
    int ans = @H_403_139@0;
    for(int i=index;i;i-=lowbit(i)) ans+=sum[i];
    return ans;
}

int update(int x,int y,int k){
    int fx=top[x],fy=top[y];
    while(fx!=fy){
        if(dep[fx]<dep[fy])swap(x,y),swap(fx,fy);
        update(tree[fx],k),update(tree[x]+@H_403_139@1,-k);
        x=fa[fx],fx=top[x];
    }
    if(dep[x]>dep[y])swap(x,y);
    update(tree[x],update(tree[y]+@H_403_139@1,-k);
}

void init(){
    cnt = @H_403_139@0;
    for(int i=@H_403_139@0;i<=n;i++)son[i]=@H_403_139@0;
}

int main(){
    while(~scanf("%d%d%d",&n,&m,&q)){
        memset(sum,@H_403_139@0,sizeof(sum));

        for(int i=@H_403_139@1;i<=n;i++) scanf("%d",&w[i]);

        for(int i=@H_403_139@1,v;i<=m;i++){
            scanf("%d%d",&u,&v);
            add(u,v);
        }

        init();
        dfs1(@H_403_139@1,@H_403_139@1);
        dfs2(@H_403_139@1,@H_403_139@1);

        for(int i=@H_403_139@1;i<=n;i++)update(tree[i],w[i]),update(tree[i]+@H_403_139@1,-w[i]);

        char ch,s[@H_403_139@10];
        int a,b,k;
        while(q--){
            scanf("%s",s);
            ch=s[@H_403_139@0];
            if(ch=='Q'){
                scanf("%d",&a);
                printf("%d\n",getSum(tree[a]));
            }
            else {
                scanf("%d%d%d",&a,&b,&k);
                if(ch=='D') k=-k;
                update(a,k);
            }
        }
        for(int i=@H_403_139@1;i<=n;i++)G[i].clear();
    }
    return @H_403_139@0;
}