AcWing 325. 計算機

題目鏈接：傳送門

要求每個點能到達的最遠點的距離
求出樹的直徑的兩個端點
對每個點和這兩個端點的距離最大值取max即是答案
當然還有一種樹形dp做法

#include <bits/stdc++.h>
#define A 20010

using namespace std;
struct node {int next, to, w;}e[A];
int head[A], num;
void add(int fr, int to, int w) {
	e[++num].next = head[fr]; e[num].to = to; e[num].w = w; head[fr] = num;
}
int n, a, b, dis[A], d1[A], d2[A]; bool vis[A];

int main(int argc, char const *argv[]) {
	while (cin >> n) {
		memset(d1, 0, sizeof d1); memset(d2, 0, sizeof d2);
		memset(head, 0, sizeof head); num = 0;
		for (int i = 2; i <= n; i++) scanf("%d%d", &a, &b), add(i, a, b), add(a, i, b);
		auto bfs = [](int s, int *dis) -> int {
			memset(dis, 0, sizeof dis); memset(vis, 0, sizeof vis);
			queue<int> q; q.push(s); vis[s] = 1;
			while (!q.empty()) {
				int fr = q.front(); q.pop();
				for (int i = head[fr]; i; i = e[i].next) {
					int ca = e[i].to;
					if (vis[ca]) continue;
					vis[ca] = 1; dis[ca] = dis[fr] + e[i].w;
					q.push(ca);
				}
			}
			int maxx = 0, pos = 0;
			for (int i = 1; i <= n; i++) if (dis[i] > maxx) maxx = dis[i], pos = i;
			return pos;
		};
		int p1 = bfs(1, dis); int p2 = bfs(p1, dis);
		bfs(p1, d1); bfs(p2, d2);
		for (int i = 1; i <= n; i++) printf("%d\n", max(d1[i], d2[i]));
	}
	return 0;
}