《洛谷CF1009F Dominant Indices》

前言：准确理解题意并转化，也是非常重要的一步。

dsu on tree裸题。

对于静态链分治的理解，更加深刻了一些，对于mx等有影响的参数。

在处理轻儿子时，也需要清空，这才是真正意义上的清空操作。

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<string,int> pii;
const int N = 1e6+5;
const int M = 1e5+5;
const LL Mod = 199999;
#define rg register
#define pi acos(-1)
#define INF 1e9
#define CT0 cin.tie(0),cout.tie(0)
#define IO ios::sync_with_stdio(false)
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
    inline LL read(){
        LL x = 0,f = 1;char c = getchar();
        while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}
        while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
        return x*f;
    }
    void print(int x){
        if(x < 0){x = -x;putchar('-');}
        if(x > 9) print(x/10);
        putchar(x%10+'0');
    }
}
using namespace FASTIO;
void FRE(){/*freopen("data1.in","r",stdin);
freopen("data1.out","w",stdout);*/}

int n,ssize[N],son[N],dep[N],Son,d[N],ans[N],mx = 0,ma;
vector<int> G[N];
void dfs(int u,int fa)
{
    ssize[u] = 1,dep[u] = dep[fa]+1;
    for(auto v : G[u])
    {
        if(v == fa) continue;
        dfs(v,u);
        ssize[u] += ssize[v];
        if(ssize[v] > ssize[son[u]]) son[u] = v;
    }
}
void slove(int u,int fa,int val)
{
    d[dep[u]] += val;
    if(d[dep[u]] > mx) mx = d[dep[u]],ma = dep[u];
    if(d[dep[u]] == mx && dep[u] < ma) ma = dep[u];
    for(auto v : G[u])
    {
        if(v == fa || v == Son) continue;
        slove(v,u,val);
    }
}
void dfs1(int u,int fa,int opt)
{
    for(auto v : G[u])
    {
        if(v == fa || v == son[u]) continue;
        dfs1(v,u,0);
    }
    if(son[u]) dfs1(son[u],u,1),Son = son[u];
    slove(u,fa,1);
    ans[u] = ma-dep[u];
    Son = 0;
    if(opt == 0) slove(u,fa,-1),mx = ma = 0;
}
int main()
{
    n = read();
    for(rg int i = 1;i < n;++i)
    {
        int x,y;x = read(),y = read();
        G[x].push_back(y);
        G[y].push_back(x);
    }
    dfs(1,0);
    dfs1(1,0,0);
    for(rg int i = 1;i <= n;++i) printf("%d\n",ans[i]);
    system("pause");
}