Aragorn’s Story

Time Limit: 20 Sec  Memory Limit: 256 MB

题目连接

http://codevs.cn/problem/1060/

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.
A的某一段完全重合，或者能够经过上下左右平移与折线A的某一段完全重合，则表示秋实大哥吹出了妹子的一部分旋律。

Input

nput 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.

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

题意

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

题解：

http://blog.csdn.net/acdreamers/article/details/10594121

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
#include <vector>

using namespace std;
const int N=50010;

int n,m,Q;
int tim;

int num[N],siz[N],top[N],son[N];
int dep[N],tid[N],ran[N],fa[N];
int head[N],to[2*N],nex[2*N],edge;

void Init()
{
    memset(head,-1,sizeof(head));
    memset(son,-1,sizeof(son));
    tim=0;
    edge=0;
}

void addedge(int u,int v)
{
    to[edge]=v,nex[edge]=head[u],head[u]=edge++;
    to[edge]=u,nex[edge]=head[v],head[v]=edge++;
}

//树链剖分部分
void dfs1(int u,int father,int d)
{
    dep[u]=d;
    fa[u]=father;
    siz[u]=1;
    for(int i=head[u];~i;i=nex[i])
    {
        int v=to[i];
        if(v!=father)
        {
            dfs1(v,u,d+1);
            siz[u]+=siz[v];
            if(son[u]==-1||siz[v]>siz[son[u]])
                son[u]=v;
        }
    }
}

void dfs2(int u,int tp)
{
    top[u]=tp;
    tid[u]=++tim;
    ran[tid[u]]=u;
    if(son[u]==-1) return;
    dfs2(son[u],tp);
    for(int i=head[u];~i;i=nex[i])
    {
        int v=to[i];
        if(v!=son[u]&&v!=fa[u])
            dfs2(v,v);
    }
}

//线段树部分
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1

int sum[4*N],col[4*N];

void PushUP(int rt)
{
    sum[rt]=max(sum[rt<<1],sum[rt<<1|1]);
}

void PushDown(int rt,int m)
{
    if(col[rt])
    {
        col[rt<<1]+=col[rt];
        col[rt<<1|1]+=col[rt];
        sum[rt<<1]+=(m-(m>>1))*col[rt];
        sum[rt<<1|1]+=(m>>1)*col[rt];
        col[rt]=0;
    }
}

void Build(int l,int r,int rt)
{
    col[rt]=0;
    if(l==r)
    {
        sum[rt]=num[ran[l]];
        return;
    }
    int mid=(l+r)>>1;
    Build(lson);
    Build(rson);
    PushUP(rt);
}

void Update(int L,int R,int v,int l,int r,int rt)
{
    if(L<=l&&R>=r)
    {
        col[rt]+=v;
        sum[rt]+=v*(r-l+1);
        return;
    }
    PushDown(rt,r-l+1);
    int mid=(l+r)>>1;
    if(L<=mid)
        Update(L,R,v,lson);
    if(R>mid)
        Update(L,R,v,rson);
    PushUP(rt);
}

int Query(int l,int r,int rt,int val)
{
    if(l==r)
        return sum[rt];
    PushDown(rt,r-l+1);
    int mid=(l+r)>>1;
    int ret=0;
    if(val<=mid) ret=Query(lson,val);
    else         ret=Query(rson,val);
    PushUP(rt);
    return ret;
}

void Change(int x,int y,int val)
{
    while(top[x]!=top[y])
    {
        if(dep[top[x]]<dep[top[y]]) swap(x,y);
        Update(tid[top[x]],tid[x],val,1,n,1);
        x=fa[top[x]];
    }
    if(dep[x]>dep[y]) swap(x,y);
    Update(tid[x],tid[y],val,1,n,1);
}

int main()
{
    char oper[5];
    int a,b,c;
    while(~scanf("%d%d%d",&n,&m,&Q))
    {
        Init();
        for(int i=1;i<=n;i++)
           scanf("%d",&num[i]);
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d",&a,&b);
            addedge(a,b);
        }
        dfs1(1,0,0);
        dfs2(1,1);
        Build(1,n,1);
        while(Q--)
        {
            scanf("%s",oper);
            if(oper[0]=='Q')
            {
                scanf("%d",&a);
                printf("%d\n",Query(1,n,1,tid[a]));
            }
            else
            {
                scanf("%d%d%d",&a,&b,&c);
                if(oper[0]=='D') c=-c;
                Change(a,b,c);
            }
        }
    }
    return 0;
}