bzoj 2561

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=2561

考虑做mst的时候，什么时候这条边不在这棵mst上呢？ 就是比他小的权值的边讲这条边的两边并进了一个联通块里面，那么对于所有的小于所求边的权值的边建一个图，然后求一个最小割使得U, V 不联通，一共做两遍加起来就是答案

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long ll;
const ll maxn = 210000;

struct node {
    ll val, size, dis, lazy; 
    node *l, *r;
}e[maxn * 8]; ll ne = 0;

void swap(node *&a, node* &b) {
    node* t = a; a = b, b = t;
}

void test(node* x) {
    if(x) {
        cout << x-> val <<" "<< x-> size << endl;
        test(x-> l), test(x-> r);
    }
}

void update(node *x) {
    x-> size = 1;
    if(x-> l) x-> size += x-> l-> size; 
    if(x-> r) x-> size += x-> r-> size;
}

void pushdown(node* x) {
    if(!x || !x-> lazy) return; 
    if(x-> l) x-> l-> val += x-> lazy, x-> l-> lazy += x-> lazy;
    if(x-> r) x-> r-> val += x-> lazy, x-> r-> lazy += x-> lazy; 
    x-> lazy = 0;
}

node* merge(node* a, node *b) {
    if(!a) return b;
    if(!b) return a;
    pushdown(a), pushdown(b);
    if(a-> val < b-> val) swap(a, b); 
    a-> r = merge(a-> r, b); 
    ll dl = a-> l ? a-> l-> size : -1; 
    ll dr = a-> r ? a-> r-> size : -1;
    if(dl < dr) swap(a-> l, a-> r); 
    a-> dis = a-> r ? a-> r-> dis + 1 : 0;
    update(a); 
    return a;
}

void pop(node* &x) {
    pushdown(x);
    x = merge(x-> l, x-> r);
}

ll n, m;

struct edge {
    ll t, d; 
    edge* next;
}se[maxn * 2], *head[maxn]; ll oe = 0;

void addedge(ll f, ll t, ll d) {
    se[oe].t = t, se[oe].d = d, se[oe].next = head[f], head[f] = se + oe ++;
}

ll int_get() {
    ll x = 0; char c = (char)getchar(); bool f = 0;
    while(!isdigit(c)) {
        if(c == '-') f = 1; 
        c = (char)getchar();
    }
    while(isdigit(c)) {
        x = x * 10 + (int)(c - '0');
        c = (char)getchar(); 
    }
    if(f) x= -x;
    return x;
}

void read() {
    n = int_get(); m = int_get();
    for(ll i = 2; i <= n; ++ i) {
        ll f, w; 
        f = int_get(), w = int_get(); 
        addedge(i, f, w), addedge(f, i, w);
    }
}

ll s[maxn], top = 0; node* rt[maxn];
ll h[maxn];

void dfs(ll x, ll fa) {
    h[x] = h[fa] + 1;
    for(edge* p = head[x]; p; p = p-> next) {
        if(p-> t != fa) dfs(p-> t, x);
    }
    s[++ top] = x;
}

ll ans[maxn];

void sov() {
    dfs(1, 0); 
    for(ll j = 1; j <= top; ++ j) {
        ll i = s[j];
        rt[i] = e + ne ++; rt[i]-> dis = 0; rt[i]-> val = 0; rt[i]-> size = 1; 
        for(edge* p = head[i]; p; p = p-> next) {
            if(h[p-> t] > h[i]) {
                if(rt[p-> t]) rt[p-> t]-> lazy += p-> d, rt[p-> t]-> val += p-> d;
                rt[i] = merge(rt[i], rt[p-> t]); 
            }
        }
        while(rt[i] && rt[i]-> val > m) pop(rt[i]); 
        ans[i] += rt[i] ? rt[i]-> size : 0;
    }
    for(ll i = 1; i <= n; ++ i) printf("%lld\n", ans[i]); 
}

int main() {
    //freopen("test.in", "r", stdin);
    //freopen("test.out", "w", stdout);
    read(), sov();
    return 0;
}