Solution to a problem P5021 [construction] track - Rockwell Valley

This question is a little tree dp meaning (Gangster light spray

I just got no idea when this question only know half the answer to this question

Why is half the answer? ? ?

topic:

Currently the construction of the track program has not been determined. Your task is to design a
Species track construction scheme, such that the track construction m in length of
the smallest maximum length of the track (i.e., track m shortest track
length as large as possible)

Minimum or maximum ...... ...... maximum usually occurs at a minimum is half the answer.

How dichotomous answer? ? ?

This question is asked of the minimum length of the maximum, it must be
half the length, that is when we can begin to set up

l = 0， r = 最大值  mid = (l + r) / 2

We obtained a length greater than equal to each track we call mid valid, and if valid track number is greater than m, then the described mid <= true answer, so we let l = mid + 1, two points continues, otherwise let r = mid.

How to transfer? ? ?

In fact, the beginning of the blind do when I have not found this to be a tree dp (escape

There is such a map, we start to consider each subtree

We assume that this node 2 subtree rooted in, from 5-2 and then this path 7 is to meet the legal conditions of the track, then we can add a direct answer, then these two edges deleted.

It may be a few remaining side.

We can see that if you use this node 2 to constitute legal length of the track, either 2 node starts to go up from 0 to Couchu legal length, either pick a side 2 below (we first assumed to be 6 to 2 sides) up to join in, can only pick a side, then we must be to pick a never used the longest side.

n = 50000, we can use to achieve an orderly lower_bound multiset side of each node and find out if you can make up one edge legal side.

This can be achieved so

int dfs(int x, int fa, int mid) {
    int len = 0;
    multiset<int> s;
    for (int i = p[x]; i != -1; i = e[i].nxt) {
        int v = e[i].v, w = e[i].w, l;
        if (v == fa) {
            continue;
        }
        l = dfs(v, x, mid) + w;
        opt[x] += opt[v];
        if (l >= mid) {
            opt[x]++;
        } else {
            s.insert(l);
        }
    }
    while (!s.empty()) {
        int now = (*s.begin());
        s.erase(s.begin());
        multiset<int>::iterator it = s.lower_bound(mid - now);
        if (it != s.end()) {
            s.erase(it);
            opt[x]++;
        } else {
            len = now;
        }
    }
    return len;
}

As Figure chrysanthemum words may get caught? ? ? I have not tried, especially if you are concerned, then you can judge it, just once and then sort lower_bound OK.

8.19 update

有同学不知道菊花图是什么

就是介个东西QWQ

菊花图通常会被卡，所以需要特判或者寻找更高效算法（ 一般是特判辣， 因为菊花图上的问题大部分比较简单的 ）

好像也没多少树上dp