codeforces 690C2 C2. Brain Network (medium)(bfs+树的直径)

最新推荐文章于 2024-08-28 22:49:31 发布

转载最新推荐文章于 2024-08-28 22:49:31 发布 · 79 阅读

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原文链接：http://www.cnblogs.com/zhangchengc919/p/5661800.html

本文介绍了一种通过两次广度优先搜索(BFS)来确定给定树的最大距离(即树的直径)的方法。此方法适用于验证僵尸大脑网络的研究假设，通过找到树中最远的两个节点间的距离来衡量僵尸的智力水平。

题目链接：

C2. Brain Network (medium)

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 uand 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

output

input

output

3

题意：

给一棵树,求树的直径;

思路：

两次bfs找树的直径,水题;

AC代码;

#include <bits/stdc++.h>
/*
#include <vector>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <cstring>
#include <algorithm>
#include <cstdio>
*/
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef  long long LL;
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e5+10;
const int maxn=1005;
const double eps=1e-10;



int n,m,vis[N],dis[N];
vector<int>ve[N];
queue<int>qu;

void bfs(int x)
{
    mst(dis,0);
    mst(vis,0);
    qu.push(x);
    vis[x]=1;
    while(!qu.empty())
    {
        int fr=qu.front();
        qu.pop();
        int len=ve[fr].size();
        for(int i=0;i<len;i++)
        {
            int y=ve[fr][i];
            if(!vis[y])
            {
                dis[y]=dis[fr]+1;
                vis[y]=1;
                qu.push(y);
            }
        }
    }
}

int main()
{
        int u,v;
        read(n);read(m);
        For(i,1,m)
        {
            read(u);read(v);
            ve[u].push_back(v);
            ve[v].push_back(u);
        }
        bfs(1);
        int s=0,ans=0;
        For(i,1,n)
        if(dis[i]>dis[s])s=i;
        bfs(s);
        For(i,1,n)ans=max(ans,dis[i]);
        cout<<ans<<"\n";

        return 0;
}

转载于:https://www.cnblogs.com/zhangchengc919/p/5661800.html