生活在树上

// Problem: 生活在树上
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/11223/D
// Memory Limit: 524288 MB
// Time Limit: 4000 ms
// 2022-03-25 20:02:25
// 
// 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 = 1e6 + 10;

int n, m;
vector<pii> g[N];
int cnt[N];
void solve() {
    cin >> n;
    for (int i = 2; i <= n ; i ++) {
        int fa, w;
        fa = read();
        w = read();
        g[i].pb({fa, w});
        g[fa].pb({i, w});
    }
    for (int i = 1; i <= n; i ++) {
        for (auto t : g[i])
            if (t.se == 1)
                cnt[i] ++;
    }
    for (int i = 1; i <= n; i ++) {
        int ans = 1;
        for (auto t : g[i])
            if (t.se == 1)
                ans += cnt[t.fi];
            else if (t.se == 2)
                ans ++;
        print(ans);
        puts("");
    }
    
}
int main () {
    // ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}

题意:给定一棵树，对于每个结点i，找出可以经过多少个不同的点，并且距离不超过2

思路:分开统计
我们先统计一个结点的儿子有多少个符合要求的，然后再针对每个i，再分开计算
如果w = 1, 那么我们就可以和ver的子节点cnt[ver]组合成2的情况
否则如果w=2，那么对总的答案贡献1