Link : https://www.acwing.com/problem/content/description/523/
Ideas :
An old and complicated subject involving a lot of knowledge points. lca + tree difference + bisection
First look at the data range is 3e5, then think about how to reduce the complexity to log (n) log (n)log（n）
What is required in the title is the shortest time required to complete m transportation plans. That is, the maximum value of the m paths is the smallest, so it is necessary to find the segment with the largest weight between the two points in the m required paths.
Then I found that this time satisfies monotonicity, and I can think of dicing the answer. Let the time be mid, in fact, the time here is the length of the path. Then for each mid value, you only need to check whether the weight between the two points can be removed to make the remaining part of the path length less than mid.
Then, among the m paths, if there are t paths whose length is greater than mid, the paths returning to 0 are included.
For each section of the path, each time it passes by +1, if a certain section of the path is added to t, then it will be deleted.
However, the time complexity of the operation of the path one at a time is too great, and it needs to be optimized. Quickly add a certain number to a certain segment at the same time and find the distance between two points. The line segment tree, tree-like array or difference is used here, and the prefix sum can be used, and each O(1) modification and query can be realized.
About the difference on the tree and the prefix sum on the tree, you can read the blog of this big guy:
https://www.luogu.com.cn/blog/Sweetlemon/s-and-d-on-a-tree
Then the difference in the tree requires the nearest common ancestor. You can use lca, tarjan or multiply, because I don't know how to tarjan so I choose multiply.
Then I can only run through a part of the points I wrote, and I don’t know what’s wrong after I changed it, orz, wait for a big guy to help debug a bug
Code :
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 400005;
const int M = 25;
#define INF 0x3f3f3f3f
int dep[maxn],pa[M][maxn];
int sum[maxn],dis[maxn],seq[maxn],cnt = 0;
int flca[maxn];
struct edge
{
int u,v,t;
void add(int a,int b,int c){
u = a;
v = b;
t = c;
}
}e[maxn];
vector<edge> g[maxn];
int n,m,a,b,c;
void dfs(int u,int fa,int d)
{
seq[++cnt] = u;
// cout<<"u"<<u<<endl;
pa[0][u] = fa;
dep[u] = d;
for(int i = 0;i < g[u].size(); i++){
int vv = g[u][i].v;
if(vv == fa) continue;
dis[vv] = dis[u] + g[u][i].t;
dfs(vv,u,d + 1);
}
}
void init()
{
dfs(1,-1,0);
for(int i = 0 ;i < M; i++){
for(int v = 1 ;v <= n; v++){
if(pa[i][v] == -1) pa[i + 1][v] = -1;
else pa[i + 1][v] = pa[i][pa[i][v]];
}
}
}
int lca(int u,int v)
{
if(dep[u] > dep[v]) swap(u,v);
for(int k = 0; k < M ;k++){
if(dep[v] - dep[u] >> k & 1)
v = pa[k][v];
}
if(u == v) return u;
for(int k = M - 1; k>= 0; k--){
if(pa[k][u] != pa[k][v]){
u = pa[k][u];
v = pa[k][v];
}
}
return pa[0][u];
}
bool check(int mid)
{
int s = 0;
memset(sum,0,sizeof(sum));
int ma = 0;
for(int i = 0;i < m; i++){
int v = e[i].v,u = e[i].u;
int l = e[i].t;
if(l > mid){
ma = max(ma,l);//路径的最大值
sum[v] += 1,sum[u] += 1;
sum[flca[i]] -= 2;
s++;
}
}
// cout<<"s"<<s<<endl;
if(!s) return true;
//dfs序
for(int i = n;i >= 1; i--){
int ne = seq[i];
sum[pa[0][i]] += sum[ne];
// cout<<"ne"<<ne<<endl;
}
for(int i = 1;i <= n; i++){
if(sum[i] == s && mid + dis[i] - dis[pa[0][i]] >= ma)
// cout<<"s"<<s<<endl;
return true;
}
return false;
}
int main()
{
// freopen("1.in","r",stdin);
// freopen("1.out","w",stdout);
memset(pa,-1,sizeof(pa));
cin>>n>>m;
for(int i = 0;i < n - 1 ; i++){
cin>>a>>b>>c;
edge e1,e2;
e1.add(a,b,c);
e2.add(b,a,c);
g[a].push_back(e1);
g[b].push_back(e2);
}
init();
int u,v,ll;
for(int i = 0;i < m ;i++){
cin>>u>>v;
flca[i] = lca(u,v);
ll = dis[v] + dis[u] - 2 * dis[flca[i]];
e[i].add(u,v,ll);
// cout<<flca[i]<<endl;
}
int l = 0,r = 1e9;
while(l < r)
{
int mid = l + r >> 1;
// cout<<mid<<endl;
if(check(mid)) r = mid;
else l = mid + 1;
}
cout<<r<<endl;
return 0;
}