[비조이 2018] 화합하기.

제목 설명:

∑x∈Path(u, v)deep(x)k ∑x∈P a t h(u, v)d e ep(x)k

제목 분석:

K가 비교적 작은 나무 단면과 배가되어 유지보수가 가능하다는 것을 알아차리다.

제목 링크:

BZOJ 5293 Luogu 4427

Ac 코드:

나무가 쪼개지는 게 느려..

#include 
#include 
#include 
#include 
#include 
const int maxm=310000;
const int mod=998244353;
int sum[maxm*4][51];
int head[maxm],to[maxm<<1],net[maxm<<1],cnt;
int deep[maxm],fa[maxm],siz[maxm],id[maxm],rank[maxm],top[maxm],son[maxm],tot;
int n,m,k;
struct node{
    int u,v,kth;
}q[maxm];
inline void addedge(int u,int v)
{
    cnt++;
    to[cnt]=v,net[cnt]=head[u],head[u]=cnt;
}
inline int read()
{
    int x=0,w=1;
    char ch=0;
    while(ch<'0'||ch>'9')
    {
        if(ch=='-') w=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
     x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
    return x*w;
}
void dfs1(int now)
{
    siz[now]=1;
    for(int i=head[now];i;i=net[i])
    if(to[i]!=fa[now])
    {
        deep[to[i]]=deep[now]+1;
        fa[to[i]]=now;
        dfs1(to[i]);
        siz[now]+=siz[to[i]];
        if(siz[son[now]]int now,int topx)
{
    rank[id[now]=++tot]=now;
    top[now]=topx;
    if(son[now]) dfs2(son[now],topx);
    for(int i=head[now];i;i=net[i])
     if(!id[to[i]]) dfs2(to[i],to[i]);
}
inline void add(int &x,int y)
{
    x+=y;
    if(x>=mod) x%=mod;
}
void build(int o,int l,int r)
{
    if(l>=r)
    {
        //printf("%d %d %d
",l,rank[l],deep[rank[l]]);
        sum[o][0]=1;
        for(int i=1;i<=k;i++)
         add(sum[o][i],(int)((1ll*sum[o][i-1]*deep[rank[l]])%mod));
        return;
    }
    int mid=(l+r)>>1;
    build((o<<1),l,mid),build((o<<1)|1,mid+1,r);
    for(int i=1;i<=k;i++)
    {
        add(sum[o][i],sum[(o<<1)][i]);
        add(sum[o][i],sum[(o<<1)|1][i]);
    }
}
int ask(int o,int l,int r,int ql,int qr,int kth)
{
    if(ql<=l&&r<=qr) return sum[o][kth];
    int mid=(l+r)>>1;
    int ans=0;
    if(ql<=mid) add(ans,ask((o<<1),l,mid,ql,qr,kth));
    if(qr>mid) add(ans,ask((o<<1)|1,mid+1,r,ql,qr,kth));
    return ans;
}
inline int treesum(int u,int v,int kth)
{
    int ans=0;
    while(top[u]!=top[v])
    {
        if(deep[top[u]]1,1,n,id[top[u]],id[u],kth));
        u=fa[top[u]];
    }
    if(deep[u]1,1,n,id[v],id[u],kth));
    return ans;
}
int main()
{
    n=read();
    for(int i=1;iint u,v;
        u=read(),v=read();
        addedge(u,v),addedge(v,u);
    }
    dfs1(1),dfs2(1,1);
    m=read();
    for(int i=1;i<=m;i++)
     q[i].u=read(),q[i].v=read(),q[i].kth=read(),k=std::max(q[i].kth,k);
    build(1,1,n);
    for(int i=1;i<=m;i++)
     printf("%d
",treesum(q[i].u,q[i].v,q[i].kth));
    return 0;
}