time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Bogdan has a birthday today and mom gave him a tree consisting of n vertecies. For every edge of the tree i, some number xi 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:
Chooses some number yi, and two vertecies ai and bi. After that, he moves along the edges of the tree from vertex ai to vertex bi 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 yi by , that is, by the result of integer division yi div xj.
Chooses some edge pi and replaces the value written in it xpi by some positive integer ci < xpi.
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 yi for each i of the first type.
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 ui, vi and xi (1 ≤ ui, vi ≤ n, ui ≠ vi, 1 ≤ xi ≤ 1018), denoting an edge that connects vertecies ui and vi, with the number xi 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 ai bi yi corresponds to a guest, who chooses the operation of the first type.
2 pi ci corresponds to a guests, who chooses the operation of the second type.
It is guaranteed that all the queries are correct, namely 1 ≤ ai, bi ≤ n, 1 ≤ pi ≤ n - 1, 1 ≤ yi ≤ 1018 and 1 ≤ ci < xpi, where xpi represents a number written on edge pi at this particular moment of time that is not necessarily equal to the initial value xpi, 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 yi through the path from ai to bi.
Examples
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
output
2
4
20
3
input
5 4
1 2 7
1 3 3
3 4 2
3 5 5
1 4 2 100
1 5 4 1
2 2 2
1 1 3 4
output
2
0
2
Note
Initially the tree looks like this:
The response to the first query is: = 2
After the third edge is changed, the tree looks like this:
The response to the second query is: = 4
In the third query the initial and final vertex coincide, that is, the answer will be the initial number 20.
After the change in the fourth edge the tree looks like this:
In the last query the answer will be: = 3
【题目链接】:http://codeforces.com/contest/593/problem/D
【题解】
两个点之间的最短路径->LCA;
两个点同时往上走;
在走的过程中除边就可以了(哪条边先除是一样的);
然后如果边的边权大于1,即最少为2；则复杂度是log2n;->nlog2n这是可以接受的;->变成0 就可以直接退出了所以是log2n;
但是就怕数据给你一大段边权全是1的情况。这样你再除一下就可能退化成O(n)了；->n^2
而我们注意到在修改边权的时候那些边权只会变小;所以最后变成1的点肯定不会再增大了；
则我们可以跳过这些点->并查集(除1的话还是不变，不管除几个1都是一样的);
LCA的话一步一步往上走就好，遇到边大于1就直接除；如果等于1就一直往上跳(并查集);
【完整代码】
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <string>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
using namespace std;
const int MAXN = 200000+100;
const int dx[5] = {0,1,-1,0,0};
const int dy[5] = {0,0,0,-1,1};
const double pi = acos(-1.0);
int n,m;
int f[MAXN],dep[MAXN];
LL w[MAXN];
vector <pair<int,int> > a[MAXN];
pair<int,int> pre[MAXN];
void rel(LL &r)
{
r = 0;
char t = getchar();
while (!isdigit(t) && t!='-') t = getchar();
LL sign = 1;
if (t == '-')sign = -1;
while (!isdigit(t)) t = getchar();
while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
r = r*sign;
}
void rei(int &r)
{
r = 0;
char t = getchar();
while (!isdigit(t)&&t!='-') t = getchar();
int sign = 1;
if (t == '-')sign = -1;
while (!isdigit(t)) t = getchar();
while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
r = r*sign;
}
void dfs(int x,int fa)
{
dep[x] = dep[fa]+1;
int len = a[x].size();
for (int i = 0;i <= len-1;i++)
{
int y = a[x][i].first,id = a[x][i].second;
if (fa == y)
continue;
pre[y] = make_pair(x,id);
dfs(y,x);
}
}
int ff(int x)
{
int id = pre[x].second;
if (w[id]!=1)
return x;
if (f[x] == x)
return f[x] = ff(pre[x].first);
return f[x] = ff(f[x]);
}
int up(int x,LL &z)
{
int prex = pre[x].first,id = pre[x].second;
if (w[id] > 1)
{
z/=w[id];
return prex;
}
return f[x] = ff(f[x]);
}
LL solve(int x,int y,LL z)
{
while (z && x!=y)
{
if (dep[x] < dep[y])
swap(x,y);
x = up(x,z);
}
return z;
}
int main()
{
rei(n);rei(m);
for (int i = 1;i <= n-1;i++)
{
int x,y;
rei(x);rei(y);rel(w[i]);
a[x].push_back(make_pair(y,i));
a[y].push_back(make_pair(x,i));
}
dfs(1,0);
for (int i = 1;i <= n;i++)
f[i] = i;
for (int i = 1;i <= m;i++)
{
int op;
rei(op);
if (op == 1)
{
int x,y;
LL z;
rei(x);rei(y);rel(z);
printf("%I64d
",solve(x,y,z));
}
else
{
int x;LL y;
rei(x);rel(y);
w[x] = y;
}
}
return 0;
}