Tree Painting 换根DP

Tree Painting

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
#include <cassert>
//#include <unordered_set>
//#include <unordered_map>
#define ll              long long
#define pii             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
#define forn(i, n)      for(int i = 0; i < int(n); i++)
using namespace std;
int dir[4][2] = { { 1,0 },{ 0,1 } ,{ 0,-1 },{ -1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-6;
const int mod = 998244353;

inline int read()
{
    int x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
inline ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}
void exgcd(ll A, ll B, ll& x, ll& y)
{
    if (B) exgcd(B, A % B, y, x), y -= A / B * x; else x = 1, y = 0;
}
inline ll qpow(ll x, ll n) {
    int r = 1;
    while (n > 0) {
        if (n & 1) r = 1ll * r * x %mod;
        n >>= 1; x = 1ll * x * x %mod;
    }
    return r;
}
inline int inv(int x) {
    return qpow(x, mod - 2);
}
ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}
/**********************************************************/
const int N = 2e5 + 5;

vector<ll> g[N];
vector<ll> size1(N, 1), f(N), a(N);
ll n;

int dfs1(int x, int fa)
{
    for (int to : g[x])
    {
        if (to != fa)
        {
            size1[x] += dfs1(to, x);
            f[x] += f[to];
        }
    }
    f[x] += size1[x];
    return size1[x];
}

ll ans;

void dfs2(int x, int fa)
{
    if (x != 1)
    {
        a[x] = a[fa] + n - size1[x] * 2ll;
        ans = max(ans, a[x]);
    }
    for (int to : g[x])
    {
        if(to!=fa)
            dfs2(to, x);
    }
}

int main()
{
    cin >> n;
    rep(i, 1, n - 1)
    {
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs1(1, -1);
    ans = a[1] = f[1];
    dfs2(1, -1);
    cout << ans << endl;
    return 0;
}