AcWing 1207. 大臣的旅费（树的直径）

Problem

这个题是一个树的最长链，也叫树的直径的模板题。

讲义一
讲义二

简单来说就是一棵带权树中求出一条路，这条路的权重之和是最大的。
我选择了一个感觉稍微简单的解法，树形DP，dist[x] 代表从x以下的子树中权重之和最大的值。

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;

class Main {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter pw = new PrintWriter(System.out);
    static int N = 100010;
    static int next[] = new int[N << 1], ver[] = new int[N << 1], head[] = new int[N << 1], edge[] = new int[N << 1];
    static int total, res;
    static int n, p, q, d;
    static boolean visit[] = new boolean[N];
    static int dist[] = new int[N];

    public static void main(String[] args) throws Exception {
        String s[] = br.readLine().split(" ");
        n = Integer.parseInt(s[0]);
        for (int i = 1; i < n; i++) {
            s = br.readLine().split(" ");
            p = Integer.parseInt(s[0]);
            q = Integer.parseInt(s[1]);
            d = Integer.parseInt(s[2]);
            add(p, q, d);
            add(q, p, d);
        }
        dp(1);
        pw.print(10 * (long) res + (long) res * (res + 1) / 2);
        pw.flush();
        pw.close();
        br.close();
    }

    public static void add(int p, int q, int d) {
        ver[++total] = q;
        edge[total] = d;
        next[total] = head[p];
        head[p] = total;
    }

    public static void dp(int x) {
        visit[x] = true;
        for (int i = head[x]; i != 0; i = next[i]) {
            int y = ver[i];
            if (visit[y]) continue;
            dp(y);
            res = Math.max(res, dist[x] + dist[y] + edge[i]);
            dist[x] = Math.max(dist[x], dist[y] + edge[i]);
        }
    }
}