educational

链接:https://codeforces.com/problemset/problem/600/E
思路:基本上算是线段树合并的裸题了吧,dfs的时候直接跟儿子合并,最后再把这个点的贡献update上去即可,然后记录答案即可。也可以用更简单的map的启发式合并,学习了一下网上的优美写法。

代码:
线段树合并

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using namespace std;

const int maxn = 2e5 + 5;
typedef long long ll;
ll maxv[maxn << 5];
int ls[maxn << 5], rs[maxn << 5], root[maxn << 5];
ll rt[maxn << 5];
int idx;
vector<int> G[maxn];
int n;
int color[maxn];
ll ans[maxn];

void (int o){
if(maxv[ls[o]] < maxv[rs[o]]){
maxv[o] = maxv[rs[o]];
rt[o] = rt[rs[o]];
}
else if(maxv[ls[o]] > maxv[rs[o]]){
maxv[o] = maxv[ls[o]];
rt[o] = rt[ls[o]];
}
else{
maxv[o] = maxv[ls[o]];
rt[o] = rt[ls[o]] + rt[rs[o]];
}
}

void update(int &o, int l, int r, int x){
if(!o) o = ++idx;
if(l == r){
maxv[o]++;
rt[o] = x;
return;
}
int mid = l + r >> 1;
if(x <= mid) update(ls[o], l, mid, x);
else update(rs[o], mid + 1, r, x);
pushup(o);
}

int merge(int o1, int o2, int l, int r){
if(!o1 || !o2) return o1 + o2;
if(l == r){
maxv[o1] += maxv[o2];
rt[o1] = l;
return o1;
}
int mid = l + r >> 1;
ls[o1] = merge(ls[o1], ls[o2], l, mid);
rs[o1] = merge(rs[o1], rs[o2], mid + 1, r);
pushup(o1);
return o1;
}

void dfs(int u, int f){
for(auto &v : G[u]){
if(v == f) continue;
dfs(v, u);
merge(root[u], root[v], 1, n);
}
update(root[u], 1, n, color[u]);
ans[u] = rt[root[u]];
}

int main(){
ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
cin >> n;
for(int i = 1; i <= n; i++) cin >> color[i], root[i] = ++idx;
for(int i = 1; i < n; i++){
int u, v;
cin >> u >> v;
G[u].emplace_back(v);
G[v].emplace_back(u);
}
dfs(1, 0);
for(int i = 1; i <= n; i++) cout << ans[i] << (i == n ? 'n' : ' ');
return 0;
}

map启发式合并

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using namespace std;

const int maxn = 1e5 + 5;
map<int, int> tr[maxn];
int n, c[maxn];
vector<int> G[maxn];
typedef long long ll;
ll ans[maxn];
ll x[maxn], y[maxn];
int id[maxn];

void merge(int &u, int v){
if(tr[u].size() < tr[v].size()) swap(u, v);
for(auto it = tr[v].begin(); it != tr[v].end(); it++){
auto &t = tr[u][it->first];
t += it->second;
if(x[u] == t) y[u] += it->first;
if(x[u] < t) x[u] = t, y[u] = it->first;
}
}

void dfs(int u, int f){
x[u] = 1, y[u] = c[u];
id[u] = u;
for(auto &v : G[u]){
if(v == f) continue;
dfs(v, u);
merge(id[u], id[v]);
}
ans[u] = y[id[u]];
}

int main(){
ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
cin >> n;
for(int i = 1; i <= n; i++) cin >> c[i];
for(int i = 1; i < n; i++){
int u, v;
cin >> u >> v;
G[u].emplace_back(v);
G[v].emplace_back(u);
}
for(int i = 1; i <= n; i++) tr[i][c[i]] = 1;
dfs(1, 0);
for(int i = 1; i <= n; i++) cout << ans[i] << (i == n ? 'n' : ' ');
return 0;
}