$Sol$
The most violent approach is to enumerate the longest side of the chain, and then counted once all chain lengths, updated $ ans $.
Here requires the largest minimum, easy to think half the answer. For half the value of $ mid $, sweeping through all the chain, if the chain length is less than equal to $ mid $, it is legal does not require treatment. Otherwise, recording all the chain the number of edges through $ 1 + $ finally find the number of passes being equal to the number of chains chain length MID $ $ maximum side, into 0, look at the longest chain is smaller than or equal MID $. $ If no such side, then $ mid $ is clearly illegal, is too small.
Maintenance times through the side of a tree difference.
$Code$
#include<bits/stdc++.h> #define il inline #define Ri register int #define go(i,a,b) for(Ri i=a;i<=b;++i) #define yes(i,a,b) for(Ri i=a;i>=b;--i) #define e(i,u) for(Ri i=b[u];i;i=a[i].nt) #define mem(a,b) memset(a,b,sizeof(a)) #define ll long long #define db double #define inf 2147483647 using namespace std; il int read() { Ri x=0,y=1;char c=getchar(); while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();} while(c>='0'&&c<='9'){x=(x<<1)+(x<<3)+c-'0';c=getchar();} return x*y; } const int N=300010; int n,m,b[N],ct,dep[N],f[N][19],dis[N],t[N],num,dd,as; bool fl; struct nd{int v,w,nt;}a[N*2]; struct nd1{int u,v,l,lc;}c[N]; il void add(Ri u,Ri v,Ri w){a[++ct]=(nd){v,w,b[u]};b[u]=ct;} il void build(Ri u,Ri fa) { dep[u]=dep[fa]+1; f[u][0]=fa; go(i,1,18)f[u][i]=f[f[u][i-1]][i-1]; e(i,u) { if(a[i].v==fa)continue; dis[a[i].v]=dis[u]+a[i].w; build(a[i].v,u); } } il int lca(Ri u,Ri v) { if(dep[u]<dep[v])swap(u,v); yes(i,18,0)if(dep[f[u][i]]>dep[v])u=f[u][i]; if(dep[u]!=dep[v])u=f[u][0]; if(u==v)return u; yes(i,18,0)if(f[u][i]!=f[v][i])u=f[u][i],v=f[v][i]; return f[u][0]; } il bool cmp(nd1 x,nd1 y){return x.l>y.l;} il int dfs(Ri u) { Ri ret=0; e(i,u) { if(a[i].v==f[u][0])continue; Ri qvq=dfs(a[i].v);ret+=qvq; if(qvq==num)fl=1,dd=max(dd,a[i].w); } ret+=t[u]; return ret; } il bool ck(Ri x) { if(c[1].l<=x)return 1; mem(t,0);num=0;dd=0;fl=0; go(i,1,m) { if(c[i].l<=x)break; Ri u=c[i].u,v=c[i].v,lc=c[i].lc; ++t[u];++t[v];t[lc]-=2;++num; } dfs(1); if(!fl)return 0; if(c[1].l-dd>x)return 0; return 1; } int main() { n=read(),m=read(); go(i,1,n-1){Ri u=read(),v=read(),w=read();add(u,v,w);add(v,u,w);} build(1,0); go(i,1,m) { Ri u=read(),v=read(),lc=lca(u,v); c[i]=(nd1){u,v,dis[u]+dis[v]-2*dis[lc],lc}; } sort(c+1,c+n+1,cmp); Ri l=0,r=c[1].l+1; while(l<=r) { Ri mid=(l+r)>>1; if(ck(mid))as=mid,r=mid-1; else l=mid+1; } printf("%d\n",as); return 0; }