P3304 [SDOI2013]直径(树的直径)

[SDOI2013]直径 - 洛谷https://www.luogu.com.cn/problem/P3304本题先跑dfs找任意一条直径，并记录这条路径和其两端端点，首先从左端点开始枚举直径上的每一个点，记录该点走其他路径的最大路径值（非本直径的路），然后开始找这些最大路径值能否等于原来的那一条路（相当于分叉出一条路能否等于原来的那么长），然后找到该点后，然后从这个点开始判断其离右端点的（与上一样）

 画的十分抽象，希望对你的理解有帮助

#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;
int ver[MAXN];
int head[MAXN];
int nxt[MAXN];
ll cost[MAXN];
int last[MAXN];
int nxxt[MAXN];
ll dis[MAXN], mmm[MAXN], op;
bool vis[MAXN];
int u, v;
int cnt;
void add(int x, int y, ll c) {
	ver[++cnt] = y;
	cost[cnt] = c;
	nxt[cnt] = head[x];
	head[x] = cnt;
}

void dfs1(int o, ll p, int q) {
	if (p > op)
		op = p, u = o;
	for (int i = head[o]; i; i = nxt[i]) {
		if (ver[i] == q)
			continue;
		dfs1(ver[i], p + cost[i], o);
	}
}

void dfs2(int o, ll p, int q) {
	last[o] = q;
	dis[o] = p;
	if (p > op)
		op = p, v = o;
	for (int i = head[o]; i; i = nxt[i]) {
		if (ver[i] == q)
			continue;
		dfs2(ver[i], p + cost[i], o);
	}
}

void dfs(int o, ll p, int q) {
	if (p > op)
		op = p;
	for (int i = head[o]; i; i = nxt[i]) {
		if (vis[ver[i]] || ver[i] == q)
			continue;
		dfs(ver[i], p + cost[i], o);
	}
}

int main() {
	scanf("%d", &n);
	int x, y;
	ll c;
	for (int i = 1; i <= n - 1; i++) {
		scanf("%d %d %lld", &x, &y, &c);
		add(x, y, c);
		add(y, x, c);
	}
	dfs1(1, 0, 0);
	op = 0;
	dfs2(u, 0, 0);
	ll distance = dis[v];
	printf("%lld\n", distance);
	for (int i = v; i; i = last[i]) {
		vis[i] = 1;
	}
	for (int i = v; i; i = last[i]) {
		op = 0;
		dfs(i, 0, 0);
		mmm[i] = op;
	}
	for (int i = last[v], j = v; i; i = last[i]) {
		nxxt[i] = j, j = i;
	}
	int tmp;
	for (int i = u; i; i = nxxt[i]) {
		if (dis[v] - dis[i] == mmm[i]) {
			tmp = i;
			break;
		}
	}
	int ans = 0;
	for (int i = tmp; i; i = last[i]) {
		if (dis[i] == mmm[i])
			break;
		ans++;
	}
	printf("%d\n", ans);
	return 0;
}