Brain Network (easy)
One particularly well-known fact about zombies is that they move and think terribly slowly. While we still don't know why their movements are so sluggish, the problem of laggy thinking has been recently resolved. It turns out that the reason is not (as previously suspected) any kind of brain defect – it's the opposite! Independent researchers confirmed that the nervous system of a zombie is highly complicated – it consists of n brains (much like a cow has several stomachs). They are interconnected by brain connectors, which are veins capable of transmitting thoughts between brains. There are two important properties such a brain network should have to function properly:
- It should be possible to exchange thoughts between any two pairs of brains (perhaps indirectly, through other brains).
- There should be no redundant brain connectors, that is, removing any brain connector would make property 1 false.
If both properties are satisfied, we say that the nervous system is valid. Unfortunately (?), if the system is not valid, the zombie stops thinking and becomes (even more) dead. Your task is to analyze a given nervous system of a zombie and find out whether it is valid.
The first line of the input contains two space-separated integers n and m (1 ≤ n, m ≤ 1000) 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, a ≠ b).
The output consists of one line, containing either yes or no depending on whether the nervous system is valid.
4 4
1 2
2 3
3 1
4 1
no
6 5
1 2
2 3
3 4
4 5
3 6
yes
分析:判断联通和环;
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],cnt;
vi a[maxn];
void dfs(int now,int pre)
{
vis[now]=;cnt++;
for(int x:a[now])
{
if(vis[x]&&x!=pre)exit(*puts("no"));
if(!vis[x])dfs(x,now);
}
}
int main()
{
int i,j,k,t;
scanf("%d%d",&n,&m);
rep(i,,m)scanf("%d%d",&j,&k),a[j].pb(k),a[k].pb(j);
dfs(j,);
if(cnt==n)puts("yes");else puts("no");
//system ("pause");
return ;
}
Brain Network (easy)的更多相关文章
- Brain Network (easy)(并查集水题)
G - Brain Network (easy) Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & ...
- 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 690C1 Brain Network (easy) (水题,判断树)
题意:给定 n 条边,判断是不是树. 析:水题,判断是不是树,首先是有没有环,这个可以用并查集来判断,然后就是边数等于顶点数减1. 代码如下: #include <bits/stdc++.h&g ...
- Brain Network (medium)
Brain Network (medium) Further research on zombie thought processes yielded interesting results. As ...
- 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 ...
- 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 ...
- 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 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
- IP相关常识
IP相关常识 一. IP地址概念 IP地址是一个32位的二进制数,它由网络ID和主机ID两部份组成,用来在网络中唯一的标识的一台计算机.网络ID用来标识计算机所处的网段:主机ID用来标识计算机在网段 ...
- Linux下将Mysql和Apache加入到系统服务里的方法
Apache加入到系统服务里面: cp /安装目录下/apache/bin/apachectl /etc/rc.d/init.d/httpd 修改httpd 在文件头部加入如下内容: ### # Co ...
- leetcode136 利用异或运算找不同的元素
Given an array of integers, every element appears twice except for one. Find that single one. Note: ...
- hdu_5727_Necklace(二分匹配)
题目连接:hdu_5727_Necklace 题意: 有2*n个珠子,n个阳珠子,n个阴珠子,现在要将这2n个珠子做成一个项链,珠子只能阴阳交替排,有些阳珠子周围如果放了指定的阴珠子就会变坏,给你一个 ...
- hdu_3555_Bomb(数位DP)
题目连接:hdu_3555_Bomb 题意:和2089一样,只是数据大了点,不过道理是一样的 #include<cstdio> #include<cstring> #defin ...
- nefu 519 昨日重现
昨日重现 Problem : 519 Time Limit : 1000ms Memory Limit : 65536K description 兴安黑熊在高中学习数学时,曾经知道这样一个公式:f(n ...
- <marquee>,视频和音频的插入,正则表达式
1.marquee 页面的自动滚动效果,不仅可以移动文字,也可以移动图片表格. <marquee></marquee>之间加内容即可. marquee的属性 1.滚动方向dir ...
- 使用Visual C++编程
1.Visual C++的主要组件 Visual C++支持遵循C++11语言标准的C++语句,C++11语言标准是在ISO/IEC 14882:2011文档中定义的.Visual C++还没有实现该 ...
- IBM Mq Spring JMS 的xml配置
<?xml version="1.0" encoding="UTF-8"?><beans xmlns="http://www.spr ...
- PHP xdebug的安装
xdebug实际上就是PHP的一个第三方扩展 安装xdebug步骤和添加一个PHP扩展一样 linux:去xdebug官网下载对应版本的源码,然后像编译其他linux扩展一样,详解我的一篇关于Linu ...