EEqual Tree Sums

原创

whiteawll 2022-08-16 14:39:35 博主文章分类：练习题目 ©著作权

文章标签 构造 dfs #define git 构造方法 文章分类 后端开发

©著作权归作者所有：来自51CTO博客作者whiteawll的原创作品，请联系作者获取转载授权，否则将追究法律责任

// Problem: E. Equal Tree Sums
// Contest: Codeforces - CodeTON Round 1 (Div. 1 + Div. 2, Rated, Prizes!)
// URL: https://codeforces.com/contest/1656/problem/E
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 2022-03-25 11:34:25
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);

inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a%mod;
        a = a * a %mod;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
inline ll inv (ll a) {
    return qmi(a, mod - 2);
}
const int N = 1e5  + 10;
int n;
int ans[N];
vi g[N];
void solve() {
    cin >> n;
    rep(i, 1, n)
        g[i].clear();
    for (int i = 1; i <= n-1 ;i ++) {
        int a, b;
        cin >> a >> b;
        g[a].pb(b);
        g[b].pb(a);
    }
    function<void(int,int,int)> cl = [&](int u, int fa, int col) {
        ans[u] = col * g[u].size();
        for (auto ver : g[u])
            if(ver != fa)
                cl(ver, u, -col);
    };
    cl(1, 1,1);
    for (int i = 1; i <= n; i ++)
        cout << ans[i] << ' ';
}
int main () {
    // ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    cin >> t;
    while (t --) solve();
    return 0;
}