【点分树】codechef Yet Another Tree Problem

已经连咕了好几天博客了；比较经典的题目

题目大意

给出一个 N 个点的树和$K_i$， 求每个点到其他所有点距离中第 $K_i$ 小的数值。

题目分析

做法一：点分树上$\log^3$

首先暴力做法：对于每个节点维护其他点距离的值域线段树。这个做法的瓶颈在于关于边$(u,v)$线段树的转移。那么可以利用点分树（为了保证复杂度）换一种容斥的思路利用重复的信息：记$f_i$为以$i$为根的点分树内所有其他点到点$i$的距离的值域线段树；$g_i$为以$i$为根的点分树内，所有点到$i$的点分树父亲的距离 的值域线段树。

询问的时候，记$LIM$为二分的长度，查询点为$pos$，$lst$的点分树父亲为$i$，那么每一层的贡献就是$f_i$中$LIM-dis(pos,i)$减去$g_{lst}$中$LIM-dis(pos,i)$的代价，注意还要特判一下$(pos,i)$这条路径是否会被算入贡献。

 #include<bits/stdc++.h>
 const int maxLog = ;
 const int maxn = ;
 const int maxm = ;
 const int maxNode = ;

 struct node
 {
     int l,r,val;
 };
 int LIM;
 struct RangeSeg
 {
     int tot,rt[maxn];
     node a[maxNode];
     void modify(int &rt, int l, int r, int pos)
     {
         if (!rt) rt = ++tot;
         ++a[rt].val;
         if (l==r) return;
         int mid = (l+r)>>;
         if (pos <= mid) modify(a[rt].l, l, mid, pos);
         else modify(a[rt].r, mid+, r, pos);
     }
     int query(int rt, int L, int R, int l, int r)
     {
         if (!rt) return ;
         if (L <= l&&r <= R) return a[rt].val;
         int mid = (l+r)>>, ret = ;
         if (L <= mid) ret = query(a[rt].l, L, R, l, mid);
         if (R > mid) ret += query(a[rt].r, L, R, mid+, r);
         return ret;
     }
     void modify(int x, int c)
     {
         modify(rt[x], , LIM, c);
     }
     int query(int x, int l, int r)
     {
         if (l <= r) return query(rt[x], l, r, , LIM);
         else return ;
     }
 }f,g;
 int n,k[maxn];
 int dep[maxn],fat[maxn][maxLog],lay[maxn];
 int size[maxn],bloTot,son[maxn],root;
 int edgeTot,head[maxn],nxt[maxm],edges[maxm];
 bool vis[maxn];

 int read()
 {
     char ch = getchar();
     int num = , fl = ;
     for (; !isdigit(ch); ch=getchar())
         if (ch=='-') fl = -;
     for (; isdigit(ch); ch=getchar())
         num = (num<<)+(num<<)+ch-;
     return num*fl;
 }
 void addedge(int u, int v)
 {
     edges[++edgeTot] = v, nxt[edgeTot] = head[u], head[u] = edgeTot;
     edges[++edgeTot] = u, nxt[edgeTot] = head[v], head[v] = edgeTot;
 }
 int lca(int x, int y)
 {
     if (dep[x] > dep[y]) std::swap(x, y);
     for (int i=; i>=; i--)
         if (dep[fat[y][i]] >= dep[x]) y = fat[y][i];
     if (x==y) return x;
     for (int i=; i>=; i--)
         if (fat[x][i]!=fat[y][i]) x = fat[x][i], y = fat[y][i];
     return fat[x][];
 }
 int dist(int x, int y){return dep[x]+dep[y]-(dep[lca(x, y)]<<);}
 void dfs(int x, int fa)
 {
     fat[x][] = fa, dep[x] = dep[fa]+;
     for (int i=; i<=; i++)
         fat[x][i] = fat[fat[x][i-]][i-];
     for (int i=head[x]; i!=-; i=nxt[i])
         if (edges[i]!=fa) dfs(edges[i], x);
 }
 void getRoot(int x, int fa)
 {
     size[x] = , son[x] = ;
     for (int i=head[x]; i!=-; i=nxt[i])
     {
         int v = edges[i];
         if (v==fa||vis[v]) continue;
         getRoot(v, x);
         size[x] += size[v];
         son[x] = std::max(son[x], size[v]);
     }
     son[x] = std::max(son[x], bloTot-size[x]);
     if (son[x] < son[root]) root = x;
 }
 void color(int Top, int x, int fa)
 {
     if (Top!=x) f.modify(Top, dist(Top, x));
     if (lay[Top]) g.modify(Top, dist(lay[Top], x));
     for (int i=head[x]; i!=-; i=nxt[i])
         if (edges[i]!=fa&&!vis[edges[i]]) color(Top, edges[i], x);
 }
 void deal(int x, int pre)
 {
     lay[x] = pre, color(x, x, x), vis[x] = true;
     for (int i=head[x]; i!=-; i=nxt[i])
     {
         int v = edges[i];
         if (vis[v]) continue;
         bloTot = size[v], root = , getRoot(v, );
         deal(root, x);
     }
 }
 int count(int x, int num)
 {
     int ret = f.query(x, , num);
     for (int i=lay[x],lst=x; i; lst=i,i=lay[i])
     {
         int d = dist(x, i);
         if (d <= num) ++ret, ret += f.query(i, , num-d)-g.query(lst, , num-d);
     }
     return ret;
 }
 int main()
 {
     memset(head, -, sizeof head);
     LIM = n = read();
     for (int i=; i<=n; i++) k[i] = n-read();
     for (int i=; i<n; i++) addedge(read(), read());
     dfs(, );
     bloTot = n, root = , son[] = n, getRoot(, );
     deal(root, );
     for (int i=; i<=n; i++)
     {
         int L = , R = LIM, ans = ;
         for (int mid=(L+R)>>; L<=R; mid=(L+R)>>)
             if (count(i, mid) < k[i]) L = mid+, ans = mid;
             else R = mid-;
         printf("%d ",ans);
     }
     return ;
 }

做法二：序列问题分块

不说了。类似的套路见#6046. 「雅礼集训 2017 Day8」爷