Codeforces Round #329 (Div. 2) D. Happy Tree Party 树链剖分

D. Happy Tree Party

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/593/problem/D

Description

Bogdan has a birthday today and mom gave him a tree consisting of n vertecies. For every edge of the tree i, some number x_i was written on it. In case you forget, a tree is a connected non-directed graph without cycles. After the present was granted, m guests consecutively come to Bogdan's party. When the i-th guest comes, he performs exactly one of the two possible operations:

1. Chooses some number y_i, and two vertecies a_i and b_i. After that, he moves along the edges of the tree from vertex a_i to vertex b_i using the shortest path (of course, such a path is unique in the tree). Every time he moves along some edge j, he replaces his current number y_i by , that is, by the result of integer division y_i div x_j.
2. Chooses some edge p_i and replaces the value written in it x_pi by some positive integer c_i < x_pi.

As Bogdan cares about his guests, he decided to ease the process. Write a program that performs all the operations requested by guests and outputs the resulting value y_i for each i of the first type.

Under two situations the player could score one point.

⋅1. If you touch a buoy before your opponent, you will get one point. For example if your opponent touch the buoy #2 before you after start, he will score one point. So when you touch the buoy #2, you won't get any point. Meanwhile, you cannot touch buoy #3 or any other buoys before touching the buoy #2.

⋅2. Ignoring the buoys and relying on dogfighting to get point. If you and your opponent meet in the same position, you can try to fight with your opponent to score one point. For the proposal of game balance, two players are not allowed to fight before buoy #2 is touched by anybody.

There are three types of players.

Speeder: As a player specializing in high speed movement, he/she tries to avoid dogfighting while attempting to gain points by touching buoys.
Fighter: As a player specializing in dogfighting, he/she always tries to fight with the opponent to score points. Since a fighter is slower than a speeder, it's difficult for him/her to score points by touching buoys when the opponent is a speeder.
All-Rounder: A balanced player between Fighter and Speeder.

There will be a training match between Asuka (All-Rounder) and Shion (Speeder).
Since the match is only a training match, the rules are simplified: the game will end after the buoy #1 is touched by anybody. Shion is a speed lover, and his strategy is very simple: touch buoy #2,#3,#4,#1 along the shortest path.

Asuka is good at dogfighting, so she will always score one point by dogfighting with Shion, and the opponent will be stunned for T seconds after dogfighting. Since Asuka is slower than Shion, she decides to fight with Shion for only one time during the match. It is also assumed that if Asuka and Shion touch the buoy in the same time, the point will be given to Asuka and Asuka could also fight with Shion at the buoy. We assume that in such scenario, the dogfighting must happen after the buoy is touched by Asuka or Shion.

The speed of Asuka is V1 m/s. The speed of Shion is V2 m/s. Is there any possibility for Asuka to win the match (to have higher score)?

Input

The first line of the input contains integers, n and m (2 ≤ n ≤ 200 000, 1 ≤ m ≤ 200 000) — the number of vertecies in the tree granted to Bogdan by his mom and the number of guests that came to the party respectively.

Next n - 1 lines contain the description of the edges. The i-th of these lines contains three integers u_i, v_i and x_i (1 ≤ u_i, v_i ≤ n, u_i ≠ v_i, 1 ≤ x_i ≤ 10¹⁸), denoting an edge that connects vertecies u_i and v_i, with the number x_i initially written on it.

The following m lines describe operations, requested by Bogdan's guests. Each description contains three or four integers and has one of the two possible forms:

• 1 a_i b_i y_i corresponds to a guest, who chooses the operation of the first type.
• 2 p_i c_i corresponds to a guests, who chooses the operation of the second type.

It is guaranteed that all the queries are correct, namely 1 ≤ a_i, b_i ≤ n, 1 ≤ p_i ≤ n - 1, 1 ≤ y_i ≤ 10¹⁸ and 1 ≤ c_i < x_pi, where x_pi represents a number written on edge p_i at this particular moment of time that is not necessarily equal to the initial value x_pi, as some decreases may have already been applied to it. The edges are numbered from 1 to n - 1 in the order they appear in the input.

Output

For each guest who chooses the operation of the first type, print the result of processing the value y_i through the path from a_i to b_i.

Sample Input

6 6
1 2 1
1 3 7
1 4 4
2 5 5
2 6 2
1 4 6 17
2 3 2
1 4 6 17
1 5 5 20
2 4 1
1 5 1 3

Sample Output

2
4
20
3

HINT

题意

 给你一颗树，然后有两个操作

1 x y z，假设x到y的链上边权分别为x1,x2,x3....，那么输出 [[[z/x1]/x2]/x3]

2 x y，把第x条边的边权改为y

题解：

有一个结论 [[[z/x1]/x2]/x3] = [z/(x1*x2*x3)]

那么我们只要维护区间乘积就好了，树链剖分裸题

但是区间乘积很容易爆LL，那么我们就用一个log来判断，如果爆了ll，那就直接把他置为1e18

然后答案直接输出0就好了

代码

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
#include <math.h>
using namespace std;
const int N=800010;
const int INF=1<<30;

int n,tim;

long long num[N];
int siz[N],top[N],son[N];
int dep[N],tid[N],Rank[N],fa[N];
int head[N],to[2*N],Next[2*N],w[2*N],edge;
int flag = 0;
struct Edge
{
    int u,v;
    long long c;
};
Edge tmp[2*N];

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

void addedge(int u,int v,int c)
{
    to[edge]=v,w[edge]=c,Next[edge]=head[u],head[u]=edge++;
    to[edge]=u,w[edge]=c,Next[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=Next[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;
    Rank[tid[u]]=u;
    if(son[u]==-1) return;
    dfs2(son[u],tp);
    for(int i=head[u]; ~i; i=Next[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
const long long kkk = 1e18+28;
long long MAX[4*N];

void PushUP(int rt)
{
    if(MAX[rt<<1]==0)
        MAX[rt]=1;
    if(MAX[rt<<1|1]==0)
        MAX[rt<<1|1]=1;
    double ppp1 = MAX[rt<<1];
    double ppp2 = MAX[rt<<1|1];
    if(log(ppp1) + log(ppp2) > log(kkk * 1.0))
        MAX[rt]=kkk+5;
    else
        MAX[rt]=MAX[rt<<1]*MAX[rt<<1|1];
}

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

void Update(int l,int r,int rt,int p,long long val)
{
    if(l==r)
    {
        MAX[rt]=val;
        return;
    }
    int mid=(l+r)>>1;
    if(p<=mid)
        Update(lson,p,val);
    else
        Update(rson,p,val);
    PushUP(rt);
}

long long Query(int l,int r,int rt,int L,int R)
{
    if(L<=l&&R>=r)
        return MAX[rt];
    int mid=(l+r)>>1;
    long long ans=1;
    if(L<=mid)
    {
        long long kkkk = Query(lson,L,R);
        if(kkkk==-1)
            flag = 1;
        if(log(ans*1.0) + log(kkkk*1.0) > log(kkk*1.0))
            ans = kkk+203;
        else
            ans*=kkkk;
    }
    if(R>mid)
    {
        long long kkkk = Query(rson,L,R);
        if(kkkk==-1)
            flag = 1;
        if(log(ans*1.0) + log(kkkk*1.0) > log(kkk*1.0))
            ans = kkk+203;
        else
            ans*=kkkk;
    }
    return ans;
}

void Change(int x,long long val)
{
    if(dep[tmp[x].u]>dep[tmp[x].v])
        Update(2,n,1,tid[tmp[x].u],val);
    else
        Update(2,n,1,tid[tmp[x].v],val);
}

long long query(int x,int y)
{
    long long ans=1;
    while(top[x]!=top[y])
    {
        if(dep[top[x]]<dep[top[y]]) swap(x,y);
        long long kkkk = Query(2,n,1,tid[top[x]],tid[x]);
        if(log(ans*1.0) + log(kkkk*1.0) > log(kkk*1.0))
            ans = kkk+203;
        else
            ans*=kkkk;
        x=fa[top[x]];
    }
    if(dep[x]>dep[y]) swap(x,y);
    if(x!=y)
    {
        long long kkkk =Query(2,n,1,tid[x]+1,tid[y]);
        if(log(ans*1.0) + log(kkkk*1.0) > log(kkk*1.0))
            ans = kkk+203;
        else
            ans*=kkkk;
    }
    return ans;
}
int MM;
int main()
{
    char oper[15];
    int a,b;
    long long c;
    Init();
    scanf("%d%d",&n,&MM);
    for(int i=1; i<n; i++)
    {
        scanf("%d%d%lld",&a,&b,&c);
        tmp[i].u=a;tmp[i].v=b;tmp[i].c=c;
        addedge(a,b,c);
    }
    dfs1(1,1,1);
    dfs2(1,1);
    //用边的孩子节点来表示该边
    for(int i=1;i<n;i++)
    {
        if(dep[tmp[i].u]>dep[tmp[i].v])
            num[tid[tmp[i].u]]=tmp[i].c;
        else
            num[tid[tmp[i].v]]=tmp[i].c;
    }
    Build(2,n,1);
    while(MM--)
    {
        scanf("%s",oper);
        if(oper[0]=='1')
        {
            long long ppp;
            scanf("%d%d",&a,&b);
            scanf("%lld",&ppp);
            flag = 0;
            long long dddd = query(a,b);
            printf("%lld
",ppp/dddd);
        }
        else
        {
            long long ppp;
            scanf("%d%lld",&a,&ppp);
            Change(a,ppp);
        }
    }
    return 0;
}