vector求数的直径

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 9;int n;
struct Edge
{int id, w;
};
vector<Edge> h[N];
int dist[N];void dfs(int u, int father, int distance)
{dist[u] = distance;for(auto node : h[u])if(node.id != father)dfs(node.id, u, distance + node.w);
}int main()
{scanf("%d", &n);for(int i = 0; i < n; i ++){int a, b, c;scanf("%d%d%d", &a, &b, &c);h[a].push_back({b, c});h[b].push_back({a, c});}dfs(1, -1, 0);int u = 1;for(int i = 1; i <= n; i ++)if(dist[i] > dist[u])u = i ;dfs(u, -1, 0);for(int i = 1; i <= n; i ++)if(dist[i] > dist[u])u = i;int s = dist[u];printf("%lld\n", s * 10 + s * (s + 1ll) / 2);return 0;
}

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链式前向星求树的直径（数组模拟邻接表）

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 9, M = 2e5 + 9;int n;
int h[N], e[M], w[M], ne[M], idx;
int dist[N];void add(int a, int b, int c)
{e[idx] = b,w[idx] = c;ne[idx] = h[a];h[a] = idx ++ ;
}void dfs(int u, int father, int distance)
{dist[u] = distance;for(int i = h[u]; ~i; i = ne[i]){int j = e[i];if(j != father)dfs(j, u, distance + w[i]);}
}int main()
{scanf("%d", &n);memset(h, -1, sizeof h);for(int i = 0; i < n; i ++){int a, b, c;scanf("%d%d%d", &a, &b, &c);add(a, b, c), add(b, a, c);}dfs(1, -1, 0);int u = 1; for(int i = 2; i <= n; i ++)if(dist[u] < dist[i])u = i;dfs(u, -1, 0);for(int i = 1; i <= n; i ++)if(dist[u] < dist[i])u = i;printf("%lld\n", dist[u] * 10 + (dist[u] + 1ll ) * dist[u] / 2);return 0;
}