codeforces 600E E. Lomsat gelral
传送门：https://codeforces.com/contest/600/problem/E
题意：
给你一颗n个节点的树，树上的每一个节点都有一种颜色，询问每一个节点所在的子树颜色数量最多的那些颜色的值的总和
题解：
维护子树颜色的数量和答案，线段树合并即可
代码：
/**
* ┏┓ ┏┓
* ┏┛┗━━━━━━━┛┗━━━┓
* ┃ ┃
* ┃ ━ ┃
* ┃ ＞ ＜ ┃
* ┃ ┃
* ┃... ⌒ ... ┃
* ┃ ┃
* ┗━┓ ┏━┛
* ┃ ┃ Code is far away from bug with the animal protecting
* ┃ ┃ 神兽保佑,代码无bug
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┗━━━┓
* ┃ ┣┓
* ┃ ┏┛
* ┗┓┓┏━┳┓┏┛
* ┃┫┫ ┃┫┫
* ┗┻┛ ┗┻┛
*/
// warm heart, wagging tail,and a smile just for you!
//
// _ooOoo_
// o8888888o
// 88" . "88
// (| -_- |)
// O = /O
// ____/`---'\____
// .' | |// `.
// / ||| : |||//
// / _||||| -:- |||||-
// | | - /// | |
// | \_| ''---/'' | |
// .-\__ `-` ___/-. /
// ___`. .' /--.-- `. . __
// ."" '< `.___\_<|>_/___.' >'"".
// | | : `- \`.;` _ /`;.`/ - ` : | |
// `-. \_ __ /__ _/ .-` / /
// ======`-.____`-.___\_____/___.-`____.-'======
// `=---='
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
// 佛祖保佑 永无BUG
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <bitset>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define bug printf("*********
")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]
"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]
"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]
"
const int maxn = 1e5 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double Pi = acos(-1);
LL gcd(LL a, LL b) {
return b ? gcd(b, a % b) : a;
}
LL lcm(LL a, LL b) {
return a / gcd(a, b) * b;
}
double dpow(double a, LL b) {
double ans = 1.0;
while(b) {
if(b % 2)ans = ans * a;
a = a * a;
b /= 2;
} return ans;
}
LL quick_pow(LL x, LL y) {
LL ans = 1;
while(y) {
if(y & 1) {
ans = ans * x % mod;
} x = x * x % mod;
y >>= 1;
} return ans;
}
struct node {
int l, r, sum, val;
LL ans;
} tree[maxn * 50];
int root[maxn];
int col[maxn];
int tree_cnt;
struct EDGE {
int v, nxt;
} edge[maxn << 1];
int head[maxn], tot;
void add_edge(int u, int v) {
edge[tot].v = v;
edge[tot].nxt = head[u];
head[u] = tot++;
}
#define ls tree[rt].l
#define rs tree[rt].r
void push_up(int rt) {
if(tree[ls].sum == tree[rs].sum) {
tree[rt].sum = tree[ls].sum;
tree[rt].val = tree[ls].sum;
tree[rt].ans = tree[ls].ans + tree[rs].ans;
} else if(tree[ls].sum > tree[rs].sum) {
tree[rt].sum = tree[ls].sum;
tree[rt].val = tree[ls].val;
tree[rt].ans = tree[ls].ans;
} else {
tree[rt].sum = tree[rs].sum;
tree[rt].val = tree[rs].val;
tree[rt].ans = tree[rs].ans;
}
}
void update(int &x, int l, int r, int pos, int val) {
if(!x) x = ++tree_cnt;
if(l == r) {
tree[x].val = l;
tree[x].sum += val;
tree[x].ans = l;
return;
}
int mid = (l + r) >> 1;
if(pos <= mid) update(tree[x].l, l, mid, pos, val);
else update(tree[x].r, mid + 1, r, pos, val);
push_up(x);
}
int merge(int x, int y, int l, int r) {
if(!x) return y;
if(!y) return x;
if(l == r) {
tree[x].val = l;
tree[x].sum += tree[y].sum;
tree[x].ans = l;
return x;
}
int mid = (l + r) >> 1;
tree[x].l = merge(tree[x].l, tree[y].l, l, mid);
tree[x].r = merge(tree[x].r, tree[y].r, mid + 1, r);
push_up(x);
return x;
}
int Max = 100000;
LL ans[maxn];
void dfs(int u, int fa) {
for(int i = head[u]; i != -1; i = edge[i].nxt) {
int v = edge[i].v;
if(v == fa) continue;
dfs(v, u);
merge(root[u], root[v], 1, Max);
}
update(root[u], 1, Max, col[u], 1);
ans[u] = tree[root[u]].ans;
}
void init() {
memset(head, -1, sizeof(head));
memset(col, 0, sizeof(col));
tree_cnt = tot = 0;
}
int main() {
#ifndef ONLINE_JUDGE
FIN
#endif
int n;
scanf("%d", &n);
init();
for(int i = 1; i <= n; i++) {
scanf("%d", &col[i]);
root[i] = i;
tree_cnt++;
}
for(int i = 1, u, v; i < n; i++) {
scanf("%d%d", &u, &v);
add_edge(u, v);
add_edge(v, u);
}
dfs(1, 0);
for(int i = 1; i <= n; i++) {
printf("%lld%c", ans[i], i == n ? '
' : ' ');
}
return 0;
}