bzoj 5293 树上差分 and 倍增

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;
}

标签：ch,anc,int,mo,5293,差分,maxn,define,bzoj
来源： https://blog.csdn.net/qq_40675883/article/details/100181023