练码力的好题.
题目链接

首先容易想到一条路径代表二维平面上的一个点.
在树上,那就求出\(dfs\)序.
我们令\(L_u=dfn[u],R_u=dfn[u]+size[u]-1\)
一条\(u\rightarrow v\)的路径可以被映射到一个点\((L_u,R_u)\)上

假设有一个盘子\(u\rightarrow v\),我们分两种情况讨论.
如果\(lca(u,v)\)不为\(u\),且不为\(v\),那么只要水果一个端点在\(u\)的子树内,另一个端点在\(v\)的子树内即可.
因此水果路径是属于\(\{(L_u,L_v),(R_u,R_v)\}\)这个矩形的.
假设\(lca(u,v)=u\),那么找到\(u\)的一个儿子在\(u\rightarrow v\)的路径上,设它为\(x\),那么只要一个点在\(v\)的子树内,一个点不在\(x\)的子树内即可.
这是两个矩形:\(\{(1,L_v),(L_x-1,R_v)\}\)和\(\{(R_x+1,L_v),(n,R_v)\}\).

那么我们把这个题转化成这样:
有若干个矩形,每次询问一个点,求覆盖它的所有矩形中权值第\(k\)小的.
我们可以整体二分.

代码如下:
还挺短
好吧其实大力压行过了\(QAQ\)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#define N (80010)
#define M (200010)
#define inf (0x7f7f7f7f)
#define rg register int
#define Label puts("NAIVE")
#define spa print(' ')
#define ent print('\n')
#define rand() (((rand())<<(16))^(rand()))
typedef long double ld;
typedef long long LL;
typedef unsigned long long ull;
using namespace std;
inline char read(){static const int IN_LEN=1000000;static char buf[IN_LEN],*s,*t;return (s==t?t=(s=buf)+fread(buf,1,IN_LEN,stdin),(s==t?-1:*s++):*s++);
}
template<class T>
inline void read(T &x){static bool iosig;static char c;for(iosig=false,c=read();!isdigit(c);c=read()){if(c=='-')iosig=true;if(c==-1)return;}for(x=0;isdigit(c);c=read())x=((x+(x<<2))<<1)+(c^'0');if(iosig)x=-x;
}
inline char readchar(){static char c;for(c=read();!isalpha(c)&&!isdigit(c);c=read())if(c==-1)return 0;return c;
}
const int OUT_LEN = 10000000;
char obuf[OUT_LEN],*ooh=obuf;
inline void print(char c) {if(ooh==obuf+OUT_LEN)fwrite(obuf,1,OUT_LEN,stdout),ooh=obuf;*ooh++=c;
}
template<class T>
inline void print(T x){static int buf[30],cnt;if(x==0)print('0');else{if(x<0)print('-'),x=-x;for(cnt=0;x;x/=10)buf[++cnt]=x%10+48;while(cnt)print((char)buf[cnt--]);}
}
inline void flush(){fwrite(obuf,1,ooh-obuf,stdout);}
struct que{int x,y,w,id;friend bool operator <(que a,que b){return a.x<b.x;}
}q[M],q1[M],q2[M];
struct chg{int ly,ry,x,w,tag,id;friend bool operator <(chg a,chg b){return a.w<b.w||a.w==b.w&&a.id<b.id;}
}a[N];
int n,m,Q,E,fi[N],ne[M],b[M],dfn[N],ind,siz[N],f[N][17],cnt,dep[N],ans[N];
struct BIT{int c[M];void add(int x,int w){for(;x<=n;x+=(x&-x))c[x]+=w;}void add(int l,int r,int w){add(l,w),add(r+1,-w);}int query(int x){int ans=0;for(;x;x-=(x&-x))ans+=c[x];return ans;}
}T;
void add(int x,int y){ne[++E]=fi[x],fi[x]=E,b[E]=y;}
void add_line(int lx,int ly,int rx,int ry,int w){if(lx>ly)swap(lx,ly),swap(rx,ry);a[++cnt]=(chg){ly,ry,lx,w,1,cnt};a[++cnt]=(chg){ly,ry,rx+1,w,-1,cnt};
}
void dfs(int u,int pre){dfn[u]=++ind,siz[u]=1,f[u][0]=pre;for(int i=fi[u];i;i=ne[i]){int v=b[i];if(v==pre)continue;dep[v]=dep[u]+1,dfs(v,u),siz[u]+=siz[v];}
}
void INIT(){dfs(1,0);for(int i=1;i<=16;i++)for(int j=1;j<=n;j++)f[j][i]=f[f[j][i-1]][i-1];
}
int lca(int x,int y){if(dep[x]<dep[y])swap(x,y);int k=dep[x]-dep[y];for(int i=16;~i;i--)if(k&(1<<i))x=f[x][i];if(x==y)return x;for(int i=16;~i;i--)if(f[x][i]!=f[y][i])x=f[x][i],y=f[y][i];if(x==y)return x;else return f[x][0];
}
int find(int x,int fa){int k=dep[x]-dep[fa]-1;for(int i=16;~i;i--)if(k&(1<<i))x=f[x][i];return x;
}
bool cmp(chg a,chg b){return a.x<b.x||a.x==b.x&&a.id<b.id;}
void solve(int L,int R,int l,int r){if(L>R)return;if(l+1>=r){for(int i=L;i<=R;i++)ans[q[i].id]=a[l].w;return;}int mid=(l+r)>>1,len=0,len1=0;if(mid&1)mid++;sort(a+l,a+mid+1,cmp);for(int i=l,j=L;i<=mid;i++){for(;a[i].x>q[j].x&&j<=R;j++){int x=T.query(q[j].y);if(x>=q[j].w)q1[++len]=q[j];else q2[++len1]=q[j],q2[len1].w-=x;}T.add(a[i].ly,a[i].ry,a[i].tag);if(i==mid)for(;j<=R;j++)q2[++len1]=q[j];}sort(a+l,a+mid+1);for(int i=1;i<=len;i++)q[L+i-1]=q1[i];for(int i=1;i<=len1;i++)q[L+len+i-1]=q2[i];solve(L,L+len-1,l,mid),solve(L+len,R,mid+1,r);
}
int main(){read(n),read(m),read(Q);for(int i=1,x,y;i<n;i++)read(x),read(y),add(x,y),add(y,x);INIT();for(int i=1,x,y,w;i<=m;i++){read(x),read(y),read(w);int rt=lca(x,y);if(rt==x||rt==y){if(rt==y)swap(x,y); int T=find(y,x);if(1<=dfn[T]-1)add_line(1,dfn[y],dfn[T]-1,dfn[y]+siz[y]-1,w);if(dfn[T]+siz[T]<=n)add_line(dfn[T]+siz[T],dfn[y],n,dfn[y]+siz[y]-1,w);}else add_line(dfn[x],dfn[y],dfn[x]+siz[x]-1,dfn[y]+siz[y]-1,w);}for(int i=1,x,y,K;i<=Q;i++){read(x),read(y),read(K);if(dfn[x]>dfn[y])swap(x,y);q[i]=(que){dfn[x],dfn[y],K,i};}sort(a+1,a+cnt+1),sort(q+1,q+Q+1),solve(1,Q,1,cnt);for(int i=1;i<=Q;i++)print(ans[i]),ent;return flush(),0;
}