poj 2763 树链剖分（单点更新，区间求值）

http://poj.org/problem?id=2763

Description

After their royal wedding, Jiajia and Wind hid away in XX Village, to enjoy their ordinary happy life. People in XX Village lived in beautiful huts. There are some pairs of huts connected by bidirectional roads. We say that huts in the same pair directly connected. XX Village is so special that we can reach any other huts starting from an arbitrary hut. If each road cannot be walked along twice, then the route between every pair is unique. 

Since Jiajia earned enough money, Wind became a housewife. Their children loved to go to other kids, then make a simple call to Wind: 'Mummy, take me home!' 

At different times, the time needed to walk along a road may be different. For example, Wind takes 5 minutes on a road normally, but may take 10 minutes if there is a lovely little dog to play with, or take 3 minutes if there is some unknown strange smell surrounding the road. 

Wind loves her children, so she would like to tell her children the exact time she will spend on the roads. Can you help her? 

Input

The first line contains three integers n, q, s. There are n huts in XX Village, q messages to process, and Wind is currently in hut s. n < 100001 , q < 100001. 

The following n-1 lines each contains three integers a, b and w. That means there is a road directly connecting hut a and b, time required is w. 1<=w<= 10000. 

The following q lines each is one of the following two types: 

Message A: 0 u 
A kid in hut u calls Wind. She should go to hut u from her current position. 
Message B: 1 i w 
The time required for i-th road is changed to w. Note that the time change will not happen when Wind is on her way. The changed can only happen when Wind is staying somewhere, waiting to take the next kid. 

Output

For each message A, print an integer X, the time required to take the next child.

Sample Input

3 3 1
1 2 1
2 3 2
0 2
1 2 3
0 3

Sample Output

1
3

/**
poj 2763  树链剖分（单点更新，区间求值）
题目大意：给定一个棵树，对于每条边的权值都在动态的变化，请求出在动态变化过程中，指定两个点的距离
解题思路：树链剖分，单点跟新，区间维护和
*/
#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
using namespace std;
const int maxn=100005;
typedef long long LL;
int fa[maxn],dep[maxn],siz[maxn],son[maxn],num[maxn];
int top[maxn],Hash[maxn],d[maxn][3];
int n,s,z,q,tree[maxn*4];
int head[maxn],ip;
void init()
{
    memset(head,-1,sizeof(head));
    memset(tree,0,sizeof(tree));
    ip=0;
}

struct note
{
    int v,w,next;
}edge[maxn*4];

void addedge(int u,int v,int w)
{
    edge[ip].v=v,edge[ip].w=w,edge[ip].next=head[u],head[u]=ip++;
}

void dfs(int u,int pre)
{
    son[u]=0,siz[u]=1;
     dep[u]=dep[pre]+1;
    fa[u]=pre;
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(v==pre)continue;
        dfs(v,u);
        siz[u]+=siz[v];
        if(siz[son[u]]<siz[v])
        {
            son[u]=v;
        }
    }
   /// printf("%d dep fa siz son %d %d %d %d\n",u,dep[u],fa[u],siz[u],son[u]);
}

void set_que(int u,int tp)
{
    num[u]=++z,top[u]=tp;
    Hash[z]=u;
    if(son[u])
    {
        set_que(son[u],tp);
    }
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(v==son[u]||v==fa[u])continue;
        set_que(v,v);
    }
}

void push_up(int root)
{
    tree[root]=tree[root<<1]+tree[root<<1|1];
}
void update(int root,int l,int r,int loc,int z)
{
    if(l>loc||r<loc) return;
    if(l==r)
    {
        tree[root]=z;
        return;
    }
    int mid=(l+r)>>1;
    update(root<<1,l,mid,loc,z);
    update(root<<1|1,mid+1,r,loc,z);
    push_up(root);
}

int query(int root,int l,int r,int x,int y)
{
    if(r<x||l>y)return 0;
    if(x<=l&&r<=y)
    {
        return tree[root];
    }
    int mid=(l+r)>>1;
    return query(root<<1,l,mid,x,y)+query(root<<1|1,mid+1,r,x,y);
}
int main()
{
    //freopen("data.txt","r",stdin);
    while(~scanf("%d%d%d",&n,&q,&s))
    {
        init();
        for(int i=1;i<n;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            d[i][0]=u,d[i][1]=v,d[i][2]=w;
            addedge(u,v,w);
            addedge(v,u,w);
        }
        int root=(n+1)>>1;
        z=0,dep[0]=0,siz[0]=0;
        dfs(root,0);
        set_que(root,root);
        for(int i=1;i<n;i++)
        {
            if(dep[d[i][0]]>dep[d[i][1]])
            {
               swap(d[i][0],d[i][1]);
            }
            update(1,1,z,num[d[i][1]],d[i][2]);
        }
        int s1;
        while(q--)
        {
            int x,y,zz;
            scanf("%d",&x);
            if(x==0)
            {
                scanf("%d",&y);
                s1=s;
                s=y;
                int f1=top[s1],f2=top[y],sum=0;
                while(f1!=f2)
                {
                    if(dep[f1]<dep[f2])
                    {
                        swap(f1,f2);
                        swap(s1,y);
                    }
                    sum+=query(1,1,z,num[f1],num[s1]);
                    s1=fa[f1],f1=top[s1];
                }
                if(y!=s1)
                {
                    if(dep[s1]>dep[y])
                    {
                        swap(s1,y);
                    }
                    sum+=query(1,1,z,num[son[s1]],num[y]);
                }
                printf("%d\n",sum);
            }
            else
            {
                scanf("%d%d",&y,&zz);
                update(1,1,z,num[d[y][1]],zz);
            }
        }
    }
    return 0;
}