DFS(深度) hihoCoder挑战赛14 B 赛车

题目传送门

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;
}