51nod树的直径图论,搜索_可以搜索到树的直径的结点-优快云博客

本文链接：https://blog.youkuaiyun.com/hesorchen/article/details/105912380

本文介绍了一种求解树的直径的高效算法，通过两次搜索找到树中最长路径。首先从任意节点开始找到最远节点A，再从A出发找到最远节点B，AB间的距离即为树的直径。文章提供了BFS和DFS两种实现方式的代码。

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题目

传送门

选择任意一个点跑到最远的点A，再以这个最远的点A为起点跑到最远的点B，AB之间的节点数就是树的直径。

原理：第一遍搜索肯定会跑到一个叶节点，从任意一个叶节点跑到最远的点，这个距离就是直径。

BFS和DFS都可以实现。

BFS代码：

struct node
{
    ll v, w, next;
} edge[200010];
ll ct = 1;
ll head[200010];
inline void add(ll u, ll v, ll w)
{
    edge[ct].v = v;
    edge[ct].w = w;
    edge[ct].next = head[u];
    head[u] = ct++;
}
ll dis[200010];
bool vis[200010];
queue<int> q;
ll n;
ll bfs(ll s)
{
    ll res;
    fill(dis, dis + n + 1, INF);
    fill(vis, vis + n + 1, 0);
    dis[s] = 0;
    q.push(s);
    while (!q.empty())
    {
        s = q.front();
        vis[s] = 1;
        q.pop();
        for (ll i = head[s]; i; i = edge[i].next)
        {
            if (vis[edge[i].v])
                continue;
            q.push(edge[i].v);
            dis[edge[i].v] = dis[s] + edge[i].w;
            res = edge[i].v;
        }
    }
    return res; //返回最远结点编号
}
int main()
{
    cin >> n;
    ll u, v, w;
    for (ll i = 1; i < n; i++)
    {
        scanf("%lld %lld", &u, &v);
        add(u, v, 1);
        add(v, u, 1);
    }
    ll s = bfs(1);
    ll pos = bfs(s);
    cout << dis[pos] << endl;
    return 0;
}

DFS代码：

struct node
{
    ll v, w, next;
} edge[200010];
ll ct = 1;
ll head[200010];
inline void add(ll u, ll v, ll w)
{
    edge[ct].v = v;
    edge[ct].w = w;
    edge[ct].next = head[u];
    head[u] = ct++;
}
ll dis[200010];
bool vis[200010];
ll n;
void dfs(ll s)
{
    for (int i = head[s]; i; i = edge[i].next)
    {
        if (vis[edge[i].v])
            continue;
        vis[edge[i].v] = 1;
        dis[edge[i].v] = dis[s] + 1;
        dfs(edge[i].v);
    }
}
int main()
{
    cin >> n;
    ll u, v, w;
    for (ll i = 1; i < n; i++)
    {
        scanf("%lld %lld", &u, &v);
        add(u, v, 1);
        add(v, u, 1);
    }
    for (int i = 1; i <= n; i++)
        dis[i] = INF, vis[i] = 0;
    dis[1] = 0;
    dfs(1);
    ll pos;
    ll maxx = 0;
    for (int i = 1; i <= n; i++)
    {
        if (dis[i] > maxx)
        {
            maxx = dis[i];
            pos = i;
        }
    }
    for (int i = 1; i <= n; i++)
        dis[i] = INF, vis[i] = 0;
    dis[pos] = 0;
    dfs(pos);
    maxx = 0;
    for (int i = 1; i <= n; i++)
        if (dis[i] > maxx)
            maxx = dis[i];
    cout << maxx << endl;
    return 0;
}