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;
}