P4408 [NOI2003] 逃学的小孩(树的直径)

[NOI2003] 逃学的小孩 - 洛谷https://www.luogu.com.cn/problem/P4408一道学习树的直径的好题。

题目要求最长的时间，说明答案应该是树的直径+max（起点离最近的朋友家的距离）

所以我们不仅仅要求树的直径，更要记录直径上的点，并遍历它们求最长延申距离。

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <cstring>
#include <set>
#include <cmath>
#include <map>
#include <bitset>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int MN = 65005;
const int MAXN = 1e6 + 10;
const int INF = 0x3f3f3f3f;
#define IOS ios::sync_with_stdio(false)
#define lowbit(x) ((x)&(-x))
using P = pair<int, int>;

int n, m;
int head[MAXN];
int nxt[MAXN];
int ver[MAXN];
ll cost[MAXN];
ll dis[MAXN];
int last[MAXN];
int cnt;
void add(int x, int y, ll c) {
	ver[++cnt] = y;
	nxt[cnt] = head[x];
	cost[cnt] = c;
	head[x] = cnt;
}
ll ans;
ll maxlen;
int recv, recu;
bool vis[MAXN];

void dfs(int p, ll rec, int fa) {
	if (maxlen < rec)
		maxlen = rec, recv = p;
	for (int i = head[p]; i; i = nxt[i]) {
		int v = ver[i];
		if (v == fa)
			continue;
//		dis[v]=dis[p]+cost[i];
		dfs(v, rec + cost[i], p);
	}
}

void dfs1(int p, int fa) {
	last[p] = fa;
	if (maxlen < dis[p])
		maxlen = dis[p], recu = p;
	for (int i = head[p]; i; i = nxt[i]) {
		int v = ver[i];
		if (v == fa)
			continue;
		dis[v] = dis[p] + cost[i];
		dfs1(v, p);
	}
}
ll mxx[MAXN];
ll mx;

void dfs2(int p, ll rec, int fa) {
	mx = max(mx, rec);
	for (int i = head[p]; i; i = nxt[i]) {
		int v = ver[i];
		if (vis[v] || v == fa)
			continue;
		dfs2(v, rec + cost[i], p);
	}
}

int main() {
	scanf("%d %d", &n, &m);
	int x, y;
	ll c;
	for (int i = 1; i <= m; i++) {
		scanf("%d %d %lld", &x, &y, &c);
		add(x, y, c);
		add(y, x, c);
	}
	dfs(1, 0, 0);
	maxlen = 0;
	dfs1(recv, 0);
	ll tmp = 0;
	for (int i = recu; i; i = last[i]) {
		vis[i] = 1;
	}
	for (int i = recu; i; i = last[i]) {
		mx = 0;
		dfs2(i, 0, 0);
		mxx[i] = mx;
	}
	mx = 0;
	for (int i = recu; i; i = last[i]) {
		tmp = min(dis[i], maxlen - dis[i]);
		mx = max(mx, mxx[i] + tmp);
	}
	printf("%lld\n", mx + maxlen);
	return 0;
}