两种方式存图 求树的直径

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;
}

---

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;
}