简单树上问题->树的直径

2次dfs solutions, 仅限边权为非负，或者没有边权（边权为1）:

void solve(){
int n;
cin >> n;

vector<vector<pair<int, int>>> al(n + 1);
for (int i = 0; i < n - 1; ++i){
int u, v, w;
cin >> u >> v >> w;
al[u].emplace_back(v, w);
al[v].emplace_back(u, w);
}

vector dist(n + 1);

function<void(int, int, int)> dfs = [&](int u, int p, int d){
dist[u] = d;
for (const auto& [v, w] : al[u]){
if (v != p){
dfs(v, u, d + w);
}
}
};

int now = 1;
dfs(now, 0, 0);

now = max_element(dist.begin(), dist.end()) - dist.begin();

dfs(now, 0, 0);

cout << *max_element(dist.begin(),dist.end()) << '\n';
}

1次dfs，树上dp（可以找到直径但是找不到路径）

void solve(){
int n;
cin >> n;

vector<vector<pair<int, int>>> al(n + 1);
for (int i = 0; i < n - 1; ++i){
int u, v, w;
cin >> u >> v >> w;
al[u].emplace_back(v, w);
al[v].emplace_back(u, w);
}

vector dp(n + 1);

int ans = 0;
function<void(int, int)> dfs = [&](int u, int p){
for (const auto& [v, w] : al[u]){
if (v != p){
dfs(v, u);
}
ans = max(ans, dp[u] + dp[v] + w);
dp[u] = max(dp[u], dp[v] + w);

}
};

dfs(1, 0);

cout << ans << '\n';
}