cf 1004E Sonya and Ice Cream_「cf1004e」sonya and ice cream

本文链接：https://blog.youkuaiyun.com/max_kibble/article/details/81008676

本文探讨了如何在有限的城市节点中找到最佳路径来开设冰激凌店，以确保所有市民能够以最短的距离到达最近的店铺。通过分析树形结构，采用滑动窗口等算法确定最优解。

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一原题

Sonya likes ice cream very much. She eats it even during programming competitions. That is why the girl decided that she wants to open her own ice cream shops.

Sonya lives in a city with $n$ junctions and $n - 1$ streets between them. All streets are two-way and connect two junctions. It is possible to travel from any junction to any other using one or more streets. City Hall allows opening shops only on junctions. The girl cannot open shops in the middle of streets.

Sonya has exactly $k$ friends whom she can trust. If she opens a shop, one of her friends has to work there and not to allow anybody to eat an ice cream not paying for it. Since Sonya does not want to skip an important competition, she will not work in shops personally.

Sonya wants all her ice cream shops to form a simple path of the length $r$ ( $1 \leq r \leq k$ ), i.e. to be located in different junctions $f_{1}, f_{2}, \dots, f_{r}$ and there is street between $f_{i}$ and $f_{i + 1}$ for each $i$ from $1$ to $r - 1$ .

The girl takes care of potential buyers, so she also wants to minimize the maximum distance between the junctions to the nearest ice cream shop. The distance between two junctions $a$ and $b$ is equal to the sum of all the street lengths that you need to pass to get from the junction $a$ to the junction $b$ . So Sonya wants to minimize

$max a min 1 \leq i \leq r d_{a, f_{i}}$

where $a$ takes a value of all possible $n$ junctions, $f_{i}$ — the junction where the $i$ -th Sonya's shop is located, and $d_{x, y}$ — the distance between the junctions $x$ and $y$ .

Sonya is not sure that she can find the optimal shops locations, that is why she is asking you to help her to open not more than $k$ shops that will form a simple path and the maximum distance between any junction and the nearest shop would be minimal.

Input

The first line contains two integers $n$ and $k$ ( $1 \leq k \leq n \leq 10^{5}$ ) — the number of junctions and friends respectively.

Each of the next $n - 1$ lines contains three integers $u_{i}$ , $v_{i}$ , and $d_{i}$ ( $1 \leq u_{i}, v_{i} \leq n$ , $v_{i} \neq u_{i}$ , $1 \leq d \leq 10^{4}$ ) — junctions that are connected by a street and the length of this street. It is guaranteed that each pair of junctions is connected by at most one street. It is guaranteed that you can get from any junctions to any other.

Output

Print one number — the minimal possible maximum distance that you need to pass to get from any junction to the nearest ice cream shop. Sonya's shops must form a simple path and the number of shops must be at most $k$ .

Note

In the first example, you can choose the path 2-4, so the answer will be 4.

$\text{[math]}$ The first example.

In the second example, you can choose the path 4-1-2, so the answer will be 7.

$\text{[math]}$ The second example.

二分析

给你一棵n个点的树，要求从树上找一条简单路径，路径上的点数不超过k，使得剩余的点到这条路径上的点的最大距离最小。比如图2，选取的点是{1, 2, 4}，剩余最远的点是7和8，距离都是7，这是最优解。

可以证明这k个点都必须在树的直径上，为直径上的每个点i求两个值：d1(i)代表i到直径左端点的距离，d2(i)代表去除直径上的点后，i为根的子树中的点到i的最大距离。最多在直径上选k个点，相当于一个长度为k的区间在直径上滑动，每个区间对应的值是：

$max\{d_1(i),diaLen-d_1(i+k-1), max\{d_2(i),d_2(i+1),...,d_2(i+k-1)\}\}$

内层的max就是一个在d2数组上滑动窗口最大值的问题。

三代码

/*
ID: maxkkibble
LANG: c++
PROB: cf 1004E
*/

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>

using namespace std;

// #define LOCAL
#define pb push_back
#define fi first
#define se second

typedef pair<int, int> PII;

const int maxn = 1e5 + 5;

int n, k;
vector<PII> g[maxn];

bool vis[maxn];  // useless util calc vector d2
int mx, id, pre[maxn], dis[maxn];

void dfs(int x, int fa) {
    pre[x] = fa;
    for (PII e: g[x]) {
        if (e.fi == fa || vis[e.fi]) continue;
        dis[e.fi] = dis[x] + e.se;
        if (dis[e.fi] > mx) mx = dis[e.fi], id = e.fi;
        dfs(e.fi, x);
    }
}

vector<int> dia, d1, d2;
int dia_len;

void calcDiameter() {
    int st, ed;
    mx = 0, id = 1;
    dfs(1, 0);
    st = id;
    mx = 0, id = st;
    dfs(st, dis[st] = 0);
    ed = id, dia_len = mx;
    while (ed != 0) {
        dia.pb(ed);
        d1.pb(dia_len - dis[ed]);
        vis[ed] = true;
        ed = pre[ed];
    }
    for (int i = 0; i < dia.size(); i++) {
        mx = 0, id = dia[i];
        dfs(dia[i], dis[dia[i]] = 0);
        d2.push_back(mx);
    }
}

int main() {
    #ifdef LOCAL
        freopen("e.in", "r", stdin);
    #endif
    ios::sync_with_stdio(false);
    cin >> n >> k;
    for (int i = 0; i < n - 1; i++) {
        int s, e, l;
        cin >> s >> e >> l;
        g[s].pb({e, l});
        g[e].pb({s, l});
    }
    
    calcDiameter();
    
    deque<PII> q;
    k = min(k, (int)dia.size());
    int ans = 1e9 + 5;
    for (int i = 0, j = 0; i <= dia.size() - k; i++) {
        while (j <= i + k - 1) {
            while (!q.empty() && d2[j] > q.back().fi)
                q.pop_back();
            q.push_back({d2[j], j});
            j++;
        }
        while (q.front().se < i) q.pop_front();
        int tmp = max({q.front().fi, d1[i], dia_len - d1[i + k -1]});
        ans = min(ans, tmp);
    }
    cout << ans << endl;
    return 0;
}