Codeforces 690C2 Brain Network (medium) (树的直径)

Brain Network (medium)

time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Further research on zombie thought processes yielded interesting results. As we know from the previous problem, the nervous system of a zombie consists of n brains and m brain connectors joining some pairs of brains together. It was observed that the intellectual abilities of a zombie depend mainly on the topology of its nervous system. More precisely, we define the distance between two brains u and v (1 ≤ u, v ≤ n) as the minimum number of brain connectors used when transmitting a thought between these two brains. The brain latency of a zombie is defined to be the maximum distance between any two of its brains. Researchers conjecture that the brain latency is the crucial parameter which determines how smart a given zombie is. Help them test this conjecture by writing a program to compute brain latencies of nervous systems.

In this problem you may assume that any nervous system given in the input is valid, i.e., it satisfies conditions (1) and (2) from the easy version.
Input

The first line of the input contains two space-separated integers n and m (1 ≤ n, m ≤ 100000) denoting the number of brains (which are conveniently numbered from 1 to n) and the number of brain connectors in the nervous system, respectively. In the next m lines, descriptions of brain connectors follow. Every connector is given as a pair of brains a b it connects (1 ≤ a, b ≤ n and a ≠ b).
Output

Print one number – the brain latency.
Examples
Input

4 3
1 2
1 3
1 4

Output

2

Input

5 4
1 2
2 3
3 4
3 5

Output

3

注： 题目一中的满足条件：1 图中不存在环 2 所有点都在图中；
思路： 题目得知该图为一棵树，所求为为树最长路径即树的直径；首先取任意点bfs求出具该点最远的点即树的直径的一段再对这个点进行bfs()所求的最远距离即为树的直径。
下面ac代码

#include<iostream>
#include<queue>
#include<cstdio>
#include<stack>
#include<string>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<set>

using namespace std;

struct node
{
    int n;
    int far;
}tmp;
int m,n;
int stu[100005];
bool vis[100005];
queue<node>Q;
vector<int>num[100005];

void bfs(int n)
{
    tmp.n = n;
    tmp.far = 0;
    Q.push(tmp);
    vis[n] = true;
    stu[n] = 0;
    while(!Q.empty())
    {
        node now = Q.front();
        Q.pop();
        vector<int>::iterator it;
        for(it = num[now.n].begin() ; it != num[now.n].end() ; ++it)
        {
            if(!vis[*it])
            {
                stu[*it] = now.far + 1;
                tmp.n = *it;
                tmp.far = stu[*it];
                Q.push(tmp);
                vis[*it] = true;
            }
        }
    }
}

int main()
{
    while(cin>>n>>m)
    {
        memset(vis , 0 ,sizeof(vis));
        memset(stu , 0 ,sizeof(stu));
        for(int i = 1 ; i <= n ; ++i)
        {
            num[i].clear();
        }
        for(int i = 1 ; i <= m ; ++i)
        {
            int st , en;
            cin>>st>>en;
            num[st].push_back(en);
            num[en].push_back(st);
        }
        bfs(1);
        int lenmax = 0;
        int weizhi = 1;
        for(int i = 1 ; i <= n ; ++i)
        {
            if(stu[i] > lenmax)
            {
                lenmax = stu[i];
                weizhi = i;
            }
        }
        memset(vis, 0 ,sizeof(vis));
        bfs(weizhi);
        lenmax = 0;
        weizhi = 1;
        for(int i = 1 ; i <= n ; ++i)
        {
            if(stu[i] > lenmax)
            {
                lenmax = stu[i];
                weizhi = i;
            }
        }
        cout<<lenmax<<endl;
    }
    return 0;
}