Codeforces 600E （dsu on tree）

傳送門

題意：

求以$i$爲根的子樹中，出現顏色次數最多的顏色編號之和。

思路：

先考慮最暴力的解法：枚舉每個節點，dfs其子樹暴力求解，複雜度$O(n^2)$
我們可以先重鏈剖分出每個節點的重兒子，處理輕鏈的答案然後合併到重兒子上，更新答案後消除輕兒子對答案的貢獻，複雜度$O(nlogn)$

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<stdio.h>
#include<string.h>
#include<queue>
#include<cmath>
#include<map>
#include<set>
#include<vector> 
using namespace std;
#define inf 0x3f3f3f3f
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define mem(a,b) memset(a,b,sizeof(a));
#define lowbit(x)  x&-x;  
#define debugint(name,x) printf("%s: %d\n",name,x);
#define debugstring(name,x) printf("%s: %s\n",name,x);
typedef long long ll;
typedef unsigned long long ull;
const double eps = 1e-6;
const int maxn = 1e5+5;
const ll mod = 1e9+7;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

vector<int>g[maxn];
ll kind[maxn],ans[maxn],n,mx,sum,col[maxn],son[maxn],sz[maxn],wson;
void dfs(int u,int fa){  //重鏈剖分 
	sz[u] = 1;
	for(auto v:g[u]){
		if(v == fa) continue;
		dfs(v,u);
		sz[u] += sz[v];
		if(sz[v] > sz[son[u]]){
			son[u] = v;
		}
	}
}
void solve(int u,int fa,int val){
	kind[col[u]] += val;
	if(kind[col[u]] > mx){
		mx = kind[col[u]];
		sum = col[u];
	}
	else if(kind[col[u]] == mx){
		sum += col[u];
	}
	for(auto v:g[u]){
		if(v == fa || v == wson) continue;
		solve(v,u,val);
	}
	
}
void dfsans(int u,int fa,int opt){ //opt表示是否要消除貢獻 
	for(auto v:g[u]){
		if(v == fa) continue;
		//dfsans(v,u,opt);
		if(v != son[u]){
			dfsans(v,u,0);  //處理輕鏈 
		}
	}
	if(son[u]){
		dfsans(son[u],u,1); //處理重鏈
		wson = son[u];
	}
	solve(u,fa,1);  //處理輕鏈的答案 
	wson = 0;
	ans[u] = sum;
	if(!opt) solve(u,fa,-1),mx = 0,sum = 0; //消除當前節點輕鏈的貢獻 
	
}
int main(){
	scanf("%d",&n);
	for(int i = 1; i <= n; i++)
		scanf("%lld",&col[i]);
	for(int i = 1; i < n; i++){
		int u,v;
		scanf("%d%d",&u,&v);
		g[u].push_back(v);
		g[v].push_back(u);
	}
	
	dfs(1,0);

	dfsans(1,0,0);
	for(int i = 1; i <= n; i++)
		printf("%lld ",ans[i]);
		
}