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  • 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
    ", ans[i]); 
    }
    
    int main() {
        //freopen("test.in", "r", stdin);
        //freopen("test.out", "w", stdout);
        read(), sov();
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/ianaesthetic/p/4251733.html
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