[Codeforces 600E] Lomsat gelral
[题目链接]
https://codeforces.com/contest/600/problem/E
[算法]
对于每一个点建立一棵线段树 , 维护当前点的子树中每个点的出现次数
可以通过深度优先遍历从下到上进行线段树合并
时间复杂度 : O(NlogN)
[代码]
#include<bits/stdc++.h> using namespace std; #define MAXN 100010 #define MAXP 5000005 typedef long long ll; typedef long double ld; typedef unsigned long long ull; struct edge { int to , nxt; } e[MAXN << 1]; int tot , n , m; int head[MAXN] , rt[MAXN] , c[MAXN]; ll ans[MAXN]; struct Segment_Tree { int sz; Segment_Tree() { sz = 0; } struct Node { int lc , rc; int mx; ll cnt; } a[MAXP]; inline void update(int x) { a[x].mx = max(a[a[x].lc].mx , a[a[x].rc].mx); if (a[a[x].lc].mx > a[a[x].rc].mx) a[x].cnt = a[a[x].lc].cnt; else if (a[a[x].rc].mx > a[a[x].lc].mx) a[x].cnt = a[a[x].rc].cnt; else a[x].cnt = a[a[x].lc].cnt + a[a[x].rc].cnt; } inline void modify(int &now , int l , int r , int x , int value) { if (!now) now = ++sz; if (l == r) { a[now].mx += value; a[now].cnt = l; return; } int mid = (l + r) >> 1; if (mid >= x) modify(a[now].lc , l , mid , x , value); else modify(a[now].rc , mid + 1 , r , x , value); update(now); } inline int merge(int x , int y , int l , int r) { if (x == 0 || y == 0) return x + y; if (l == r) { a[x].mx += a[y].mx; a[x].cnt = l; return x; } int mid = (l + r) >> 1; a[x].lc = merge(a[x].lc , a[y].lc , l , mid); a[x].rc = merge(a[x].rc , a[y].rc , mid + 1 , r); update(x); return x; } inline ll getans(int u) { return a[rt[u]].cnt; } } SGT; template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); } template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); } template <typename T> inline void read(T &x) { T f = 1; x = 0; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0'; x *= f; } inline void addedge(int u , int v) { ++tot; e[tot] = (edge){v , head[u]}; head[u] = tot; } inline void calc(int u , int father) { for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (v == father) continue; calc(v , u); rt[u] = SGT.merge(rt[u] , rt[v] , 1 , n); } ans[u] = SGT.getans(u); } int main() { read(n); for (int i = 1; i <= n; i++) { read(c[i]); SGT.modify(rt[i] , 1 , n , c[i] , 1); } for (int i = 1; i < n; i++) { int x , y; read(x); read(y); addedge(x , y); addedge(y , x); } calc(1 , 0); for (int i = 1; i <= n; i++) printf("%lld " , ans[i]); printf("\n"); return 0; }