CodeForces 593D Happy Tree Party(树链剖分（边权） or LCA+并查集)

题意：给你一棵数，n个点n-1条边，每条边有个权值，q次询问，询问有两种：

1 a b y : a到b的路径上不断进行y = y/xi（像下取整），问最后y的值

2 p c :  将第p条边的权值改为c，c一定比原来这条边的值小

思路：

树剖：线段树记录区间的乘积，查询的时候当两者重链的top相等的时候注意去掉top点的权值。
每条边的权值都是在1e18内，所以如果线段树某个元素代表的权值之乘积大于1e18的话就直接标记成0。

并查集+LCA也能做，过段时间再补。

代码：

#include<bits/stdc++.h>using namespace std;typedef long long ll;const int maxn = 1e6+5;const ll INF = 1e18;int n, m, tot, k;int fa[maxn], tree[maxn], son[maxn], deep[maxn], num[maxn], top[maxn], edge_id[maxn];ll treeFac[maxn];struct node{    int u, v;    ll w;}Edge[maxn];vector<int> g[maxn];void init(){    tot = k = 0;    memset(son, 0, sizeof(son));    memset(num, 0, sizeof(num));    for(int i = 0; i <= n; i++)        g[i].clear();}void dfs1(int u, int pre, int d){    fa[u] = pre;    deep[u] = d;    num[u] = 1;    for(int i = 0; i < g[u].size(); i++)    {        int v = g[u][i];        if(v == pre) continue;        dfs1(v, u, d+1);        num[u] += num[v];        if(!son[u] || num[v] > num[son[u]])            son[u] = v;    }}void dfs2(int u, int tp){    tree[u] = ++tot;    top[u] = tp;    if(!son[u]) return ;    dfs2(son[u], tp);    for(int i = 0; i < g[u].size(); i++)    {        int v = g[u][i];        if(v != fa[u] && v != son[u])            dfs2(v, v);    }}ll mul(ll a, ll b){    if(!a || !b) return 0;    if(INF/a < b) return 0;    return a*b;}void pushup(int root){    treeFac[root] = mul(treeFac[root*2], treeFac[root*2+1]);}void build(int root, int l, int r){    treeFac[root] = 1;    if(l == r) return ;    int mid = (l+r)/2;    build(root*2, l, mid);    build(root*2+1, mid+1, r);}void update(int root, int l, int r, int pos, ll val){    if(l == r)    {        treeFac[root] = val;        return ;    }    int mid = (l+r)/2;    if(pos <= mid) update(root*2, l, mid, pos, val);    else update(root*2+1, mid+1, r, pos, val);    pushup(root);}ll query(int root, int l, int r, int i, int j){    if(i <= l && j >= r)        return treeFac[root];    int mid = (l+r)/2;    ll res = 1;    if(i <= mid) res = mul(res, query(root*2, l, mid, i, j));    if(j > mid) res = mul(res, query(root*2+1, mid+1, r, i, j));    return res;}ll cal(int u, int v){    int f1 = top[u], f2 = top[v];    ll res = 1;    while(f1 != f2)    {        if(deep[f1] < deep[f2]) swap(f1, f2), swap(u, v);        res = mul(res, query(1, 1, n, tree[f1], tree[u]));        u = fa[f1];        f1 = top[u];    }    if(u == v) return res;    if(deep[u] > deep[v]) swap(u, v);    res = mul(res, query(1, 1, n, tree[son[u]], tree[v]));    return res;}int main(void){    while(cin >> n >> m)    {        init();        int u, v;        ll w;        for(int i = 1; i < n; i++)        {            scanf("%d%d%I64d", &u, &v, &w);            g[u].push_back(v);            g[v].push_back(u);            Edge[i].u = u, Edge[i].v = v, Edge[i].w = w;        }        dfs1(1, 0, 0);        dfs2(1, 1);        build(1, 1, n);        for(int i = 1; i < n; i++)        {            if(deep[Edge[i].u] > deep[Edge[i].v])                edge_id[i] = Edge[i].u;            else                edge_id[i] = Edge[i].v;        }        for(int i = 1; i < n; i++)            update(1, 1, n, tree[edge_id[i]], Edge[i].w);        while(m--)        {            int cmd;            scanf("%d", &cmd);            if(cmd == 1)            {                scanf("%d%d%I64d", &u, &v, &w);                ll ans = cal(u, v);                if(!ans) puts("0");                else printf("%I64d\n", w/ans);            }            else            {                scanf("%d%I64d", &u, &w);                update(1, 1, n, tree[edge_id[u]], w);            }        }    }    return 0;}