题意
给出一个树,求出每个节点的子树中出现次数最多的颜色的编号和
Sol
dsu on tree的裸题。
一会儿好好总结总结qwq
#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int MAXN = 1e5 + ;
inline int read() {
char c = getchar(); int x = , f = ;
while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
while(c >= '' && c <= '') x = x * + c - '', c = getchar();
return x * f;
}
int N, col[MAXN], son[MAXN], siz[MAXN], cnt[MAXN], Mx, Son;
LL sum = , ans[MAXN];
vector<int> v[MAXN];
void dfs(int x, int fa) {
siz[x] = ;
for(int i = ; i < v[x].size(); i++) {
int to = v[x][i];
if(to == fa) continue;
dfs(to, x);
siz[x] += siz[to];
if(siz[to] > siz[son[x]]) son[x] = to;
}
}
void add(int x, int fa, int val) {
cnt[col[x]] += val;
if(cnt[col[x]] > Mx) Mx = cnt[col[x]], sum = col[x];
else if(cnt[col[x]] == Mx) sum += (LL)col[x];
for(int i = ; i < v[x].size(); i++) {
int to = v[x][i];
if(to == fa || to == Son) continue;
add(to, x, val);
}
}
void dfs2(int x, int fa, int opt) {
for(int i = ; i < v[x].size(); i++) {
int to = v[x][i];
if(to == fa) continue;
if(to != son[x]) dfs2(to, x, );
}
if(son[x]) dfs2(son[x], x, ), Son = son[x]; add(x, fa, ); Son = ;
ans[x] = sum;
if(!opt) add(x, fa, -), sum = , Mx = ;
}
int main() {
N = read();
for(int i = ; i <= N; i++) col[i] = read();
for(int i = ; i <= N - ; i++) {
int x = read(), y = read();
v[x].push_back(y); v[y].push_back(x);
}
dfs(, );
dfs2(, , );
for(int i = ; i <= N; i++) printf("%I64d ", ans[i]);
return ;
}