题目链接

题意:就是给你一颗这样的树,用一个$y$来除以两点之间每条边的权值,比如$3->7$,问最后的y的是多少,修改操作是把权值变成更小的。

这个$(y<=10^{18})$除的权值如果是$>=2$,那么最多除60几次就变成0了,问题关键是路径上会有好多1存在,这时候我们可以用并查集把他们并到一块,这样就能跳着查了。

具体查法:

从$u$到$LCA(u,v)$,路径上除一遍。

从$v$到$LCA(u,v)$,路径上除一遍。

修改操作如果变成1,就与前面的点合并。

 #include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std; typedef long long ll;
const int maxn = 2e5 + ;
///加边
int cnt = , h[maxn];
struct edge
{
int to, pre;
ll v;
} e[maxn << ];
void add(int from, int to, ll v)
{
cnt++;
e[cnt].pre = h[from]; ///5-->3-->1-->0
e[cnt].to = to;
e[cnt].v = v;
h[from] = cnt;
}
///LCA
int dist[maxn];
int dep[maxn];
int anc[maxn][]; ///2分的父亲节点
ll len[maxn];
void dfs(int u, int fa)
{
for(int i = h[u]; i; i = e[i].pre)
{
int v = e[i].to;
if(v == fa) continue;
dist[v] = dist[u] + e[i].v;
dep[v] = dep[u] + ;
anc[v][] = u;
len[v] = e[i].v;
dfs(v, u);
}
}
void LCA_init(int n)
{
for(int j = ; ( << j) < n; j++)
for(int i = ; i <= n; i++) if(anc[i][j-])
anc[i][j] = anc[anc[i][j-]][j-];
}
int LCA(int u, int v)
{
int log;
if(dep[u] < dep[v]) swap(u, v);
for(log = ; ( << log) < dep[u]; log++);
for(int i = log; i >= ; i--)
if(dep[u] - ( << i) >= dep[v]) u = anc[u][i];
if(u == v) return u;
for(int i = log; i >= ; i--)
if(anc[u][i] && anc[u][i] != anc[v][i])
u = anc[u][i], v = anc[v][i];
return anc[u][];
}
int fa[maxn];
int Find(int x)
{
if(x == fa[x]) return x;
return fa[x] = Find(fa[x]);
}
int main()
{
int n, m; scanf("%d %d", &n, &m);
cnt = ;
for(int i = ; i <= n; i++) h[i] = ;
for(int i = ; i <= n - ; i++)
{
int u, v;
ll w; scanf("%d %d %lld", &u, &v, &w);
add(u, v, w), add(v, u, w);
}
///LCA初始化
dfs(, );
LCA_init(n);
///并查集初始化
for(int i = ; i <= n; i++) fa[i] = i;
while(m--)
{
int op, a, b, p;
ll c, y;
scanf("%d", &op);
if(op == )
{
scanf("%d %d %lld", &a, &b, &y);
int fp = LCA(a, b);
a = Find(a);
while(dep[a] > dep[fp] && y > 0LL)
{
if(len[a] == 1LL)
{
fa[a] = anc[a][];
a = Find(a);
}
else
{
y /= len[a];
a = Find(anc[a][]);
}
}
b = Find(b);
while(dep[b] > dep[fp] && y > 0LL)
{
if(len[b] == 1LL)
{
fa[b] = anc[b][];
b = Find(b);
}
else
{
y /= len[b];
b = Find(anc[b][]);
}
}
printf("%lld\n", y);
}
else
{
scanf("%d %lld", &p, &c);
int u = e[p * ].to, v = e[(p * ) ^ ].to;
if(u == anc[v][])
{
len[v] = c;
if(c == 1LL)
{
fa[v] = u;
}
}
else
{
len[u] = c;
if(c == 1LL)
{
fa[u] = v;
}
}
}
}
return ;
}

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