题意：问你树上两点之间路径上 sigma(每个点深度^k);
解：
1、树上倍增时维护每个深度k次幂到根的前缀和
2、树上差分+简单容斥
#include<bits/stdc++.h> using namespace std; #define MAXN 300010 #define maxn 300010 #define en '\n' #define ll long long int MAXLOG; const int mo=998244353; const int INF = 1e8; const int inf = 1e8; int T,tot = 1; int dep[maxn],anc[maxn][66],n,m; int val[66]; int a[maxn][66]; template<class T>void rep(T &x){x%=mo;x+=mo;x%=mo;} template<class T>void rd(T &x) { x=0;int f=0;char ch=getchar(); while(ch<'0'||ch>'9') {f|=(ch=='-');ch=getchar();} while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();} x=f?-x:x; return; } struct node{ int nxt,v; }edge[maxn<<1]; int head[maxn]; void add(int u,int v){ edge[++tot].nxt=head[u];head[u]=tot;edge[tot].v=v; } void dfs(int x,int fa){ dep[x]=dep[fa]+1; val[0]=1; for(int i=1;i<=50;i++){ val[i]=((ll)val[i-1]*dep[x])%mo; } for(int i=1;i<=50;i++){ a[x][i]=(a[fa][i]+val[i])%mo; } anc[x][0]=fa; if(!anc[x][0])anc[x][0]=1; for(int i=1;i<MAXLOG;i++){ anc[x][i]=anc[anc[x][i-1]][i-1]; } for(int i=head[x];i;i=edge[i].nxt){ int y=edge[i].v; if(y==fa)continue; dfs(y,x); } } int lca(int x,int y){ if(dep[x]>dep[y])swap(x,y); int t=dep[y]-dep[x]; for(int i=0;i<MAXLOG;i++){ if((t>>i)&1){ y=anc[y][i]; } } if(y==x)return x; for(int i=MAXLOG-1;i>=0;--i){ if(anc[x][i]==anc[y][i])continue; x=anc[x][i],y=anc[y][i]; } return anc[x][0]; } signed main() { #ifdef local freopen("input2.txt","r",stdin); #endif // local cin>>n; MAXLOG=log10(1.0*n)/log10(2)+2; for(int i=1;i<n;i++){ int x,y; rd(x),rd(y); add(x,y),add(y,x); } dep[0]=-1; dfs(1,0); cin>>m; for(int i=0;i<m;i++){ int x,y,k; rd(x),rd(y),rd(k); int LCA=lca(x,y); ll tem=(ll)a[x][k]+a[y][k]-a[LCA][k]-a[anc[LCA][0]][k]; rep(tem); cout<<tem<<en; } return 0; }来源：51CTO
作者：fnq9999
链接：https://blog.csdn.net/qq_40675883/article/details/100181023