标签:ble continue https ++ template 算法 sum 模板 temp
题意:
求子树众数。
思路:
\(dsu\ on\ tree\)模板题,用一个桶记录即可。
感觉\(dsu\ on\ tree\)这个算法的涉及真是巧妙呀,保留重链的信息,不断暴力轻边,并且不断在子树内递归下去。又由于轻边数量不会超过\(O(logn)\),所以总的时间复杂度控制在\(O(nlogn)\)。
/*
* Author: heyuhhh
* Created Time: 2019/11/13 15:09:25
*/
#include <bits/stdc++.h>
#define MP make_pair
#define fi first
#define se second
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
#define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
void err() { std::cout << '\n'; }
template<typename T, typename...Args>
void err(T a, Args...args) { std::cout << a << ' '; err(args...); }
#else
#define dbg(...)
#endif
void pt() {std::cout << '\n'; }
template<typename T, typename...Args>
void pt(T a, Args...args) {std::cout << a << ' '; pt(args...); }
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 1e5 + 5;
int n;
int c[N];
struct Edge{
int v, next;
}e[N << 1];
int head[N], tot;
void adde(int u, int v) {
e[tot].v = v; e[tot].next = head[u]; head[u] = tot++;
}
int Son[N], sz[N];
void dfs(int u, int fa) {
int mx = 0, son = 0;
sz[u] = 1;
for(int i = head[u]; i != -1; i = e[i].next) {
int v = e[i].v;
if(v == fa) continue;
dfs(v, u);
sz[u] += sz[v];
if(sz[v] > mx) {
mx = sz[v], son = v;
}
}
Son[u] = son;
}
ll sum, ans[N];
int son, Max;
int cnt[N];
void add(int u, int fa, int val) {
cnt[c[u]] += val;
if(cnt[c[u]] > Max) Max = cnt[c[u]], sum = c[u];
else if(cnt[c[u]] == Max) sum += c[u];
for(int i = head[u]; i != -1; i = e[i].next) {
int v = e[i].v;
if(v == fa || v == son) continue;
add(v, u, val);
}
}
void dfs2(int u, int fa, int op) {
for(int i = head[u]; i != -1; i = e[i].next) {
int v = e[i].v;
if(v == fa || v == Son[u]) continue;
dfs2(v, u, 0);
}
if(Son[u]) dfs2(Son[u], u, 1);
son = Son[u];
add(u, fa, 1);
ans[u] = sum;
son = 0;
if(!op) add(u, fa, -1), Max = sum = 0;
}
void run(){
memset(head, -1, sizeof(head));
for(int i = 1; i <= n; i++) cin >> c[i];
for(int i = 1; i < n; i++) {
int u, v; cin >> u >> v;
adde(u, v); adde(v, u);
}
dfs(1, 0);
dfs2(1, 0, 1);
for(int i = 1; i <= n; i++) cout << ans[i] << " \n"[i == n];
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
while(cin >> n) run();
return 0;
}
【cf600】E. Lomsat gelral(树上启发式合并)
标签:ble continue https ++ template 算法 sum 模板 temp
原文地址:https://www.cnblogs.com/heyuhhh/p/11862681.html