CodeForces 690C2 Brain Network (medium)（树上DP）

题意：给定一棵树中，让你计算它的直径，也就是两点间的最大距离。

析：就是一个树上DP，用两次BFS或都一次DFS就可以搞定。但两次的时间是一样的。

代码如下：

#include<bits/stdc++.h>

using namespace std;
const int maxn = 1e5 + 5;
vector<int> G[maxn];
int f[maxn], g[maxn], l[maxn];

int dfs(int root, int fa){
    if(f[root] != -1)  return f[root];
    if(!G[root].size())  return f[root] = 0;
    int ans = root, m = 0, mm = 0;
    for(int i = 0; i < G[root].size(); ++i){
        int u = G[root][i];
        if(u == fa)   continue;
        if(dfs(u, root) + 1 > m){
            m = f[u] + 1;
            ans = u;
        }
    }
    l[root] = ans;
    for(int i = 0; i < G[root].size(); ++i){
        int u = G[root][i];
        if(f[u] + 1 > mm && u != l[root])  mm = f[u] + 1;
    }
    g[root] = mm;
    return f[root] =  m;
}

int main(){
    int n, m, u, v;
    cin >> n >> m;
    while(m--){
        scanf("%d %d", &u, &v);
        G[u].push_back(v);
        G[v].push_back(u);
    }

    memset(f, -1, sizeof(f));
    int ans = 0;
    for(int i = 1; i <= n; ++i)
        if(f[i] == -1)  dfs(i, -1);
    for(int i = 1; i <= n; ++i)   ans = max(ans, f[i]+g[i]);
    cout << ans << endl;
    return 0;
}

 两次BFS：

#include<bits/stdc++.h>

using namespace std;
const int maxn = 1e5 + 5;
int d[maxn];
vector<int> G[maxn];
int vis[maxn], vvis[maxn];

int bfs(int root){
    memset(vis, 0, sizeof(vis));
    memset(d, 0, sizeof(d));
    vis[root] = 1;  vvis[root] = 1;
    queue<int> q; q.push(root);
    int ans = root, mmax = 0;
    while(!q.empty()){
        root = q.front();  q.pop();
        for(int i = 0; i < G[root].size(); ++i){
            int u = G[root][i];
            if(vis[u])  continue;
            q.push(u);
            vis[u] = vvis[u] = 1;
            d[u] = d[root] + 1;
            if(mmax < d[u]){
                mmax = d[u];
                ans = u;
            }
        }
    }
    return ans;
}

int solve(int root){
    int u = bfs(root);
    int v = bfs(u);
    return d[v];
}

int main(){
    int n, m, u, v;
    cin >> n >> m;
    for(int i = 0; i < m; ++i){
        scanf("%d %d", &u, &v);
        G[u].push_back(v);
        G[v].push_back(u);
    }

    memset(vvis, 0, sizeof(vvis));
    int ans = 0;
    for(int i = 1; i <= n; ++i)
        if(!vvis[i])   ans = max(ans, solve(i));
    cout << ans << endl;
    return 0;
}