题目链接

题目

题目描述

In ICPCCamp there were n towns conveniently numbered with \(1, 2, \dots, n\)
connected with (n - 1) roads.
The i-th road connecting towns aia_iai and bib_ibi has length \(c_i\) .
It is guaranteed that any two cities reach each other using only roads.

Bobo would like to build (n - 1) highways so that any two towns reach each using only highways.
Building a highway between towns x and y costs him \(\delta(x, y)\) cents,
where \(\delta(x, y)\) is the length of the shortest path between towns x and y using roads.

As Bobo is rich, he would like to find the most expensive way to build the (n - 1) highways.

输入描述

The input contains zero or more test cases and is terminated by end-of-file. For each test case:

The first line contains an integer n.
The i-th of the following (n - 1) lines contains three integers aia_iai, bib_ibi and cic_ici.

• \(1 \leq n \leq 10^5\)
• \(1 \leq a_i, b_i \leq n\)
• \(1 \leq c_i \leq 10^8\)
• The number of test cases does not exceed 10.

输出描述

For each test case, output an integer which denotes the result.

示例1

输入

5
1 2 2
1 3 1
2 4 2
3 5 1
5
1 2 2
1 4 1
3 4 1
4 5 2

输出

19
15

题解

知识点：DFS，树的直径。

重修 \(n-1\) 条路，使得花费和最大，每条路的花费是两端原先的最短路。

容易证明，只要以树的直径两端为道路的一端连接其他点，即可最大化花费。因此，先通过两次dfs找到树的直径端点，然后遍历每个点取距离最远的一端即可。

时间复杂度 \(O(n+m)\)

空间复杂度 \(O(n+m)\)

代码

#include <bits/stdc++.h>
#define ll long long

using namespace std;

template<class T>
struct Graph {
    struct edge {
        int v, nxt;
        T w;
    };
    int idx;
    vector<int> h;
    vector<edge> e;

    Graph(int n, int m) :idx(0), h(n + 1), e(m + 1) {}

    void clear(int n, int m) {
        idx = 0;
        h.assign(n + 1, 0);
        e.assign(m + 1, { 0,0,0 });
    }

    void add(int u, int v, T w) {
        e[++idx] = edge{ v,h[u],w };
        h[u] = idx;
    }
};

int main() {
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n;
    while (cin >> n) {
        Graph<int> g(n, n << 1);
        for (int i = 2;i <= n;i++) {
            int u, v, w;
            cin >> u >> v >> w;
            g.add(u, v, w);
            g.add(v, u, w);
        }

        function<void(int, int, vector<ll> &)> dfs = [&](int u, int fa, vector<ll> &dis) {
            for (int i = g.h[u];i;i = g.e[i].nxt) {
                int v = g.e[i].v, w = g.e[i].w;
                if (v == fa) continue;
                dis[v] = dis[u] + w;
                dfs(v, u, dis);
            }
        };

        vector<ll> dis(n + 1);
        dfs(1, 0, dis);
        int pos1 = 0;
        for (int i = 1;i <= n;i++)
            if (dis[i] > dis[pos1]) pos1 = i;

        vector<ll> dis1(n + 1);
        dfs(pos1, 0, dis1);
        int pos2 = 0;
        for (int i = 1;i <= n;i++)
            if (dis1[i] > dis1[pos2]) pos2 = i;

        vector<ll> dis2(n + 1);
        dfs(pos2, 0, dis2);
        ll ans = dis1[pos2];
        for (int i = 1;i <= n;i++) {
            if (i == pos1 || i == pos2) continue;
            ans += max(dis1[i], dis2[i]);
        }
        cout << ans << '\n';
    }

    return 0;
}