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.
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).
Print one number – the brain latency.
4 3
1 2
1 3
1 4
2
5 4
1 2
2 3
3 4
3 5
3
分析:树中最长路,两次bfs;
代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include <ext/rope>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define vi vector<int>
#define pii pair<int,int>
#define mod 1000000007
#define inf 0x3f3f3f3f
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
const int maxn=1e5+;
const int dis[][]={{,},{-,},{,-},{,}};
using namespace std;
using namespace __gnu_cxx;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}
int n,m,vis[maxn],ma,now;
vi a[maxn];
int bfs(int x)
{
memset(vis,,sizeof(vis));ma=;
queue<int>p;p.push(x);vis[x]=;
while(!p.empty())
{
int u=p.front();p.pop();
for(int y:a[u])
{
if(!vis[y])
{
vis[y]=vis[u]+;
p.push(y);
if(ma<vis[y])ma=vis[y],now=y;
}
}
}
return now;
}
int main()
{
int i,j,k,t;
scanf("%d%d",&n,&m);
while(m--)
{
scanf("%d%d",&j,&k);
a[j].pb(k),a[k].pb(j);
}
j=bfs(bfs());
printf("%d\n",ma-);
//system ("pause");
return ;
}
Brain Network (medium)的更多相关文章
- Brain Network (medium)(DFS)
H - Brain Network (medium) Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d &am ...
- codeforces 690C2 C2. Brain Network (medium)(bfs+树的直径)
题目链接: C2. Brain Network (medium) time limit per test 2 seconds memory limit per test 256 megabytes i ...
- CodeForces 690C2 Brain Network (medium)(树上DP)
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/std ...
- Brain Network (easy)
Brain Network (easy) One particularly well-known fact about zombies is that they move and think terr ...
- CF 690C3. Brain Network (hard) from Helvetic Coding Contest 2016 online mirror (teams, unrated)
题目描述 Brain Network (hard) 这个问题就是给出一个不断加边的树,保证每一次加边之后都只有一个连通块(每一次连的点都是之前出现过的),问每一次加边之后树的直径. 算法 每一次增加一 ...
- Brain Network (easy)(并查集水题)
G - Brain Network (easy) Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & ...
- codeforces 690C3 C3. Brain Network (hard)(lca)
题目链接: C3. Brain Network (hard) time limit per test 2 seconds memory limit per test 256 megabytes inp ...
- codeforces 690C1 C1. Brain Network (easy)(水题)
题目链接: C1. Brain Network (easy) time limit per test 2 seconds memory limit per test 256 megabytes inp ...
- Codeforces 690 C3. Brain Network (hard) LCA
C3. Brain Network (hard) Breaking news from zombie neurology! It turns out that – contrary to prev ...
随机推荐
- html5学习(二)音频audio
音频格式 当前,audio 元素支持三种音频格式: IE 9 Firefox 3.5 Opera 10.5 Chrome 3.0 Safari 3.0 Ogg Vorbis √ √ √ M ...
- 已有打开的与此 Command 相关联的 DataReader,必须首先将它关闭
已有打开的与此 Command 相关联的 DataReader,必须首先将它关闭 引用: http://www.cnblogs.com/maxao/archive/2011/03/18/19881 ...
- java编码问题
工作中经常遇到java编码问题,由于缺乏研究,总是无法给出确切的答案,这个周末在网上查了一些资料,在此做些汇总. 问题一:在java中读取文件时应该采用什么编码? Java读取文件的方式总体可以分为两 ...
- 轻松创建nodejs服务器(1):一个简单nodejs服务器例子
这篇文章主要介绍了一个简单nodejs服务器例子,本文实现了一个简单的hello world例子,并展示如何运行这个服务器,需要的朋友可以参考下 我们先来实现一个简单的例子,hello world ...
- oracle即时客户端安装方法
http://blog.csdn.net/magicboylinw/article/details/7025885 Oracle Instant Client(即时客户端) 安装与配置 oracleO ...
- NET:交换机的背板带宽,交换容量,包转发率区别
交换机的背板带宽,交换容量,包转发率区别 背板带宽指的是背板整个的交换容量,交换容量指cpu的交换容量,包转发指的是三层转发的容量 一.背板带宽 1.交换机背板带宽含义 交换机的背板带宽也叫背板容量, ...
- zf-关于交换工具配置文件,交换的“列名字段”前面加个“0,1,2”的意思。
- 【stack】模拟网页浏览 poj 1028
#include<stdio.h> #include<string.h> int main() { ][]; ]; int i,depth; strcpy(s[]," ...
- Coupons and Discounts
Coupons and Discounts time limit per test 1 second memory limit per test 256 megabytes input standar ...
- Dense Subsequence
Dense Subsequence time limit per test 2 seconds memory limit per test 256 megabytes input standard i ...