传送门我会让你知道哪里有题面吗（逃

思路

显然不能模拟苹果下掉的过程，考虑计算每个苹果对询问的贡献。

显然一开始就有的苹果可以看做第0天变出来的，于是只需要考虑变出来的苹果了。

设当前询问节点\(x\)，时间为\(t_1\)天早上。

设考虑的苹果一开始在\(t_2\)晚上从\(y\)节点上变了出来。

那么造成贡献的条件就是

1. \[ t_2<t_1 \]

2. \[ dfn_x\leq dfn_y \leq low_x \]

3. \[ dep_y+t_2\geq dep_x +t_1-1 \]

显然这是个三维偏序，裸上CDQ分治即可。

注意同一天早晚的区别，按时间sort时要小心。

代码

#include<bits/stdc++.h>
clock_t t=clock();
namespace my_std{using namespace std;#define pii pair<int,int>#define fir first#define sec second#define MP make_pair#define rep(i,x,y) for (int i=(x);i<=(y);i++)#define drep(i,x,y) for (int i=(x);i>=(y);i--)#define go(x) for (int i=head[x];i;i=edge[i].nxt)#define templ template<typename T>#define sz 101010typedef long long ll;typedef double db;mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}templ inline void read(T& t){t=0;char f=0,ch=getchar();double d=0.1;while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}t=(f?-t:t);}template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}char sr[1<<21],z[20];int C=-1,Z=0;inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}inline void print(register int x){if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;while(z[++Z]=x%10+48,x/=10);while(sr[++C]=z[Z],--Z);sr[++C]='\n';}void file(){#ifndef ONLINE_JUDGEfreopen("a.in","r",stdin);#endif}inline void chktime(){#ifndef ONLINE_JUDGEcout<<(clock()-t)/1000.0<<'\n';#endif}#ifdef modll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}ll inv(ll x){return ksm(x,mod-2);}#elsell ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}#endif
//  inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;int n,m,Q,M;
ll a[sz];
struct hh{int t,nxt;}edge[sz<<1];
int head[sz],ecnt;
void make_edge(int f,int t)
{edge[++ecnt]=(hh){t,head[f]};head[f]=ecnt;edge[++ecnt]=(hh){f,head[t]};head[t]=ecnt;
}int dfn[sz],low[sz],dep[sz],T;
void dfs(int x,int fa)
{dfn[x]=++T;dep[x]=dep[fa]+1;go(x) if(edge[i].t!=fa) dfs(edge[i].t,x);low[x]=T;
}struct hhh
{int type; // 0:modify,1:queryint p,d;ll w; // dfn:p,dep:d,+=wint l,r,D,id; // dfn:[l,r],dep:\geq Dint ti; // time
}q[sz<<2];
inline bool cmp(const hhh &x,const hhh &y)
{if (x.type!=y.type) return x.type;if (x.type==0) return x.d>y.d;return x.D>y.D;
}
inline bool cmpt(const hhh &x,const hhh &y)
{if (x.type==y.type) return x.ti<y.ti;else if (x.type) return x.ti<=y.ti;return x.ti<y.ti;
}
ll ans[sz];ll tr[sz];
void add(int x,ll w){while (x<=n) tr[x]+=w,x+=(x&(-x));}
ll query(int x){ll ret=0;while (x) ret+=tr[x],x-=(x&(-x));return ret;}void solve(int l,int r)
{if (l==r) return;int mid=(l+r)>>1;solve(l,mid);solve(mid+1,r);int p=l;rep(i,mid+1,r) if (q[i].type){while (p<=mid){if (q[p].type) ++p;else if (q[p].d>=q[i].D) add(q[p].p,q[p].w),++p;else break;}ans[q[i].id]+=query(q[i].r)-query(q[i].l-1);}rep(i,l,p-1) if (!q[i].type) add(q[i].p,-q[i].w);sort(q+l,q+r+1,cmp);
}int main()
{file();read(n,m,Q);rep(i,1,n) read(a[i]);int x,y,z;rep(i,1,n-1) read(x,y),make_edge(x,y);dfs(1,0);rep(i,1,n) q[++M]=(hhh){0,dfn[i],dep[i],a[i],0,0,0,0,0};while (m--) read(x,y,z),q[++M]=(hhh){0,dfn[y],dep[y]+x,z,0,0,0,0,x};rep(i,1,Q) read(z,x),q[++M]=(hhh){1,0,0,0,dfn[x],low[x],dep[x]+z-1,i,z};sort(q+1,q+M+1,cmpt);solve(1,M);rep(i,1,Q) printf("%lld\n",ans[i]);return 0;
}