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题意：中文题面
分析：放官方题解，就是从1为根节点深搜记录节点的深度，选出最大的深度的点，将该到达该点的节点都vis掉，然后再重新计算没有vis的点的深度，找最大的相加就是答案。放张图好理解：
收获：计算树的节点的深度
代码：
/************************************************
* Author :Running_Time
* Created Time :2015-8-31 14:03:09
* File Name :B.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int dep[N];
bool vis[N];
vector<int> G[N];
int fa[N];
void DFS(int u, int d) {
dep[u] = d;
for (int v, i=0; i<G[u].size (); ++i) {
v = G[u][i];
if (vis[v]) continue;
fa[v] = u;
DFS (v, d + 1);
}
}
int main(void) {
int n; scanf ("%d", &n);
for (int u, v, i=1; i<n; ++i) {
scanf ("%d%d", &u, &v);
G[u].push_back (v);
}
fa[1] = 0;
DFS (1, 0);
int mx = 0, id = 0;
for (int i=1; i<=n; ++i) {
if (mx < dep[i]) {
mx = dep[i]; id = i;
}
}
int ans = mx;
while (id) {
vis[id] = true;
id = fa[id];
}
for (int i=1; i<=n; ++i) {
if (vis[i]) {
DFS (i, 0);
}
}
mx = 0;
for (int i=1; i<=n; ++i) {
if (!vis[i]) mx = max (mx, dep[i]);
}
printf ("%d\n", ans + mx);
return 0;
}
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原文地址：http://www.cnblogs.com/Running-Time/p/4773012.html
