分析

考虑直接用倍增跳会TLE，设\(f[x][i]\)表示以\(x\)为起点每次跳\(i\)步的点权和，

这可以预处理出来，综合一下两种做法，当\(i>\sqrt{n}\)时直接上倍增，否则预处理\(f\)即可

如果用长链剖分求树上\(k\)级祖先那么就可以去掉\(\log\)

代码

#include

#include

#define rr register

using namespace std;

const int N=50011,M=224;

typedef long long lll; lll g[N][M];

struct node{int y,next;}e[N<<1];

int f[N][16],n,LCA[N],a[N],b[N],p[N],as[N],et=1,F[M],dep[N];

inline signed iut(){

rr int ans=0,f=1; rr char c=getchar();

while (!isdigit(c)) f=(c=='-')?-f:f,c=getchar();

while (isdigit(c)) ans=(ans<<3)+(ans<<1)+(c^48),c=getchar();

return ans*f;

}

inline void print(lll ans){

if (ans>9) print(ans/10);

putchar(ans%10+48);

}

inline void dfs(int x,int fa){

f[x][0]=fa,F[0]=x,dep[x]=dep[fa]+1;

for (rr int i=1;i<16&&f[x][i-1];++i)

f[x][i]=f[f[x][i-1]][i-1];

for (rr int i=1;i

for (rr int i=1;i

for (rr int i=as[x];i;i=e[i].next)

if (e[i].y!=fa) dfs(e[i].y,x);

}

inline signed lca(int x,int y){

if (dep[x]

for (rr int i=15;~i;--i)

if (dep[f[x][i]]>=dep[y]) x=f[x][i];

if (x==y) return x;

for (rr int i=15;~i;--i)

if (f[x][i]!=f[y][i])

x=f[x][i],y=f[y][i];

return f[x][0];

}

inline signed Get(int x,int y){

for (rr int i=15;~i;--i)

if ((y>>i)&1) x=f[x][i];

return x;

}

inline lll answ(int x,int LCA,int y,int P){

rr lll ans=0;

if (P>=M){

ans+=a[x];

for (;dep[x]-P>=dep[LCA];)

ans+=a[x=Get(x,P)];

rr int now=dep[x]-dep[LCA];

if (now&&P-now>dep[y]-dep[LCA]) ans+=a[y];

else{

x=Get(y,dep[y]-dep[LCA]-P*(now>0)+now);

if (x!=LCA) ans+=a[x];

for (;dep[x]+P<=dep[y];)

x=Get(y,dep[y]-dep[x]-P),ans+=a[x];

if (x!=y) ans+=a[y];

}

}else{

rr int step=(dep[x]-dep[LCA])/P;

ans+=g[x][P]-g[Get(x,(step+1)*P)][P];

rr int now=dep[x]-dep[LCA]-P*step;

if (now&&P-now>dep[y]-dep[LCA]) ans+=a[y];

else{

x=Get(y,dep[y]-dep[LCA]-P*(y!=LCA)+now);

step=(dep[y]-dep[x])/P;

rr int t=Get(y,dep[y]-dep[x]-step*P);

if (t!=LCA) ans+=g[t][P]-g[Get(x,P)][P];

if (t!=y) ans+=a[y];

}

}

return ans;

}

signed main(){

n=iut();

for (rr int i=1;i<=n;++i) a[i]=iut();

for (rr int i=1;i

rr int x=iut(),y=iut();

e[++et]=(node){y,as[x]},as[x]=et;

e[++et]=(node){x,as[y]},as[y]=et;

}

dfs(1,0);

for (rr int i=1;i<=n;++i) b[i]=iut();

for (rr int i=1;i

for (rr int i=1;i

return 0;

}