cf1009F. Dominant Indices

题意：

1号节点为根，问对于以每个点为根的子树种，求某个深度节点最多的层数。如果多个相等，输出深度小的。

题解：

直接dsu就完事了，相当于是求子树的众数，但是注意常数，一开始因为常数过大在第100点wa了
将可以合并的dfs套在一起写，减少常数
好像线段树合并也可以做，但我还不会emm

代码：

// Problem: F. Dominant Indices
// Contest: Codeforces - Educational Codeforces Round 47 (Rated for Div. 2)
// URL: https://codeforces.com/contest/1009/problem/F
// Memory Limit: 512 MB
// Time Limit: 4500 ms
// Data:2021-09-03 11:43:05
// By Jozky#include <bits/stdc++.h>
#include <unordered_map>
#define debug(a, b) printf("%s = %d\n", a, b);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
clock_t startTime, endTime;
//Fe~Jozky
const ll INF_ll= 1e18;
const int INF_int= 0x3f3f3f3f;
void read(){};
template <typename _Tp, typename... _Tps> void read(_Tp& x, _Tps&... Ar)
{x= 0;char c= getchar();bool flag= 0;while (c < '0' || c > '9')flag|= (c == '-'), c= getchar();while (c >= '0' && c <= '9')x= (x << 3) + (x << 1) + (c ^ 48), c= getchar();if (flag)x= -x;read(Ar...);
}
template <typename T> inline void write(T x)
{if (x < 0) {x= ~(x - 1);putchar('-');}if (x > 9)write(x / 10);putchar(x % 10 + '0');
}
void rd_test()
{
#ifdef LOCALstartTime= clock();freopen("in.txt", "r", stdin);
#endif
}
void Time_test()
{
#ifdef LOCALendTime= clock();printf("\nRun Time:%lfs\n", (double)(endTime - startTime) / CLOCKS_PER_SEC);
#endif
}
int n;
const int maxn= 2e6 + 9;
vector<int> vec[maxn];
int dep[maxn];
int siz[maxn];
int son[maxn];
int Son;
void dfs(int u, int fa)
{dep[u]= dep[fa] + 1;siz[u]= 1;for (auto v : vec[u]) {if (v == fa)continue;dfs(v, u);siz[u]+= siz[v];if (siz[v] > siz[son[u]])son[u]= v;}
}
int ans[maxn];
int num[maxn];
int Ans, Num;
void cal(int u, int fa, int val)
{num[dep[u]]+= val;if (val == 1) {if (num[dep[u]] > Num || (num[dep[u]] == Num && dep[u] < Ans)) {Ans= dep[u];Num= num[dep[u]];}}for (auto v : vec[u]) {if (v == fa || v == Son)continue;cal(v, u, val);}
}
// PII solve(int u, int fa, int rt)
// {
// PII maxx;
// maxx.first= num[dep[u]];
// maxx.second= dep[u] - dep[rt];
// for (auto v : vec[u]) {
// if (v == fa)
// continue;
// PII ans= solve(v, u, rt);
// if (ans.first > maxx.first) {
// maxx.first= ans.first;
// maxx.second= ans.second;
// }
// if (ans.first == maxx.first) {
// maxx.second= min(maxx.second, ans.second);
// }
// }
// return maxx;
// }
void dfs2(int u, int fa, int keep)
{for (auto v : vec[u]) {if (v == fa || v == son[u])continue;dfs2(v, u, 0);}if (son[u]) {dfs2(son[u], u, 1);Son= son[u];}cal(u, fa, 1);// PII p= solve(u, fa, u);ans[u]= Ans - dep[u];Son= 0;if (!keep) {cal(u, fa, -1);Num= 0;Ans= 0;}
}
int main()
{//rd_test();read(n);for (int i= 1; i < n; i++) {int x, y;read(x, y);vec[x].push_back(y);vec[y].push_back(x);}dfs(1, 0);dfs2(1, 0, 0);for (int i= 1; i <= n; i++)printf("%d\n", ans[i]);//Time_test();
}