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
- linux的pvtrace环境配置
1.查看当前ubuntu版本号 froid@ubuntu:~/Desktop$ lsb_release -aNo LSB modules are available.Distributor ID: ...
- 查询页面checkbox使用
HTML <input type="checkbox" id="IsChildGroup" name="IsChildGroup" v ...
- Word试卷文档模型化解析存储到数据库
最近在搞一套在线的考试系统,有许多人反映试题的新增比较麻烦(需要逐个输入),于是呼就整个了试卷批量导入了 poi实现word转html 模型化解析html html转Map数组 Map数组(数组的操作 ...
- cmstop框架中的js设计content.js
控制cmstop框架中action的js 内容模块 找出当前页面的js的思路01先找显示页面的当前文件.在页面文件中-->找(编辑,删除)按钮-->找获取这个按钮的js选择器 02看加载的 ...
- Java学习笔记之[ 利用扫描仪Scanner进行数据输入 ]
/*********数据的输入********//**利用扫描仪Scanner进行数据输入 怎么使用扫描仪Scanner *1.放在类声明之前,引入扫描仪 import java.util.Scann ...
- 利用文本编辑器输入课堂上练习的Hello.java,并在JDK环境下编译和运行。
- Ubuntu 网管服务器配置
1.设置Linux内核支持ip数据包的转发 echo "1" > /proc/sys/net/ipv4/ip_forward or vi /etc/sysctl.conf ...
- 未能加载文件或程序集“ICSharpCode.SharpZipLib, Version=0.86.0.518, Culture=neutral, PublicKeyToken=1b03e6acf1164f73”或它的某一个依赖项
未能加载文件或程序集“ICSharpCode.SharpZipLib, Version=0.86.0.518, Culture=neutral, PublicKeyToken=1b03e6acf116 ...
- MySQL的数据类型【总结】
1.时间类型 MySQL的DateTime,TimeStamp,Date和Time数据类型. DATETIME类型用在你需要同时包含日期和时间信息的值时.MySQL检索并且以'YYYY-MM-DD H ...
- maven之pom
记录一下最近的pom的相关设置,plugin的官方地址配置:http://maven.apache.org/plugins/index.html 看了网上说了很多例子,有很多不清楚,看一下官方的,会有 ...