题意：中文题。
析：首先二分是很容易想出来的，然后主要是判断这个解合不合法，先二分答案 mid，因为有 m 个计划，所以只要添加虫洞的肯定是所有的时间长于 mid 的计划 中，也就是是那些的共同边，这个就可以用树上差分来做了，假设 s 到 t，那么让in[s]++，in[t]++，in[lca(s, t)] -= 2，其中in 表示的是 该结点与其父结点的边的计数，最后再跑一次dfs，把所有的权值都累加上去，这样就能知道哪些是共同的边了。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 3e5 + 20;
const int maxm = 100 + 10;
const ULL mod = 10007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
struct Edge{
int to, val, next;
};
Edge edge[maxn<<1];
int head[maxn], cnt;
inline void addEdge(int u, int v, int val){
edge[cnt].to = v;
edge[cnt].val = val;
edge[cnt].next = head[u];
head[u] = cnt++;
}
int p[21][maxn];
int dep[maxn], dp[maxn], dist[maxn];
struct Road{
int u, v, lca, dist;
};
Road road[maxn];
inline void dfs(int u, int fa, int d){
p[0][u] = fa;
dep[u] = d;
for(int i = head[u]; ~i; i = edge[i].next){
int v = edge[i].to;
if(v == fa) continue;
dp[v] = dp[u] + edge[i].val;
dist[v] = edge[i].val;
dfs(v, u, d + 1);
}
}
void init(){
ms(p, -1); dfs(1, -1, 0);
for(int k = 0; k < 20; ++k)
for(int u = 1; u <= n; ++u)
if(p[k][u] > 0) p[k+1][u] = p[k][p[k][u]];
}
inline int LCA(int u, int v){
if(dep[u] > dep[v]) swap(u, v);
for(int k = 0; k < 20; ++k)
if(dep[v] - dep[u] >> k & 1) v = p[k][v];
if(u == v) return u;
for(int k = 19; k >= 0; --k)
if(p[k][u] != p[k][v]){
u = p[k][u];
v = p[k][v];
}
return p[0][u];
}
int in[maxn];
inline void dfs1(int u, int fa){
for(int i = head[u]; ~i; i = edge[i].next){
int v = edge[i].to;
if(v == fa) continue;
dfs1(v, u); in[u] += in[v];
}
}
inline bool judge(int mid){
int ans = 0, cnt = 0;
for(int i = 1; i <= n; ++i) in[i] = 0;
for(int i = 0; i < m; ++i) if(road[i].dist > mid){
ans = max(ans, road[i].dist - mid);
++cnt;
++in[road[i].u];
++in[road[i].v];
in[road[i].lca] -= 2;
}
if(cnt == 0) return true;
dfs1(1, -1);
for(int i = 1; i <= n; ++i) if(in[i] == cnt && dist[i] >= ans) return true;
return false;
}
int main(){
scanf("%d %d", &n, &m);
ms(head, -1); cnt = 0;
for(int i = 1; i < n; ++i){
int u, v, c;
scanf("%d %d %d", &u, &v, &c);
addEdge(u, v, c);
addEdge(v, u, c);
}
init();
int l = 0, r = 0;
for(int i = 0; i < m; ++i){
scanf("%d %d", &road[i].u, &road[i].v);
road[i].lca = LCA(road[i].u, road[i].v);
road[i].dist = dp[road[i].u] + dp[road[i].v] - (dp[road[i].lca]<<1);
r = max(r, road[i].dist);
}
while(l <= r){
int m = l + r >> 1;
if(judge(m)) r = m - 1;
else l = m + 1;
}
printf("%d\n", l);
return 0;
}