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:

  1. It should be possible to exchange thoughts between any two pairs of brains (perhaps indirectly, through other brains).
  2. 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.

Input

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 ab it connects (1 ≤ a, b ≤ na ≠ b).

Output

The output consists of one line, containing either yes or no depending on whether the nervous system is valid.

Examples
input
4 4
1 2
2 3
3 1
4 1
output
no
input
6 5
1 2
2 3
3 4
4 5
3 6
output
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

  1. Brain Network (easy)(并查集水题)

    G - Brain Network (easy) Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & ...

  2. 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 ...

  3. 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 ...

  4. Brain Network (medium)

    Brain Network (medium) Further research on zombie thought processes yielded interesting results. As ...

  5. Brain Network (medium)(DFS)

    H - Brain Network (medium) Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d &am ...

  6. 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 ...

  7. 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 ...

  8. 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 ...

  9. 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

  1. android 线程池的使用

    转自http://www.trinea.cn/android/java-android-thread-pool/ Java(Android)线程池 介绍new Thread的弊端及Java四种线程池的 ...

  2. iOS TextField输入框点击键盘时随着键盘上移

    -(void)textFieldDidBeginEditing:(UITextField *)textField { CGRect frame = textField.frame; int offse ...

  3. VMware+Windbg双机调试

    虚拟机使用XP系统:

  4. java 线程池 并行 执行

    https://github.com/donaldlee2008/JerryMultiThread/blob/master/src/com/jerry/threadpool/ThreadPoolTes ...

  5. Struts2原理图

  6. raise()函数

    kill和raise函数用来发送信号, 区别在于: kill把信号发送给进程或进程组. kill(pid_t pid, int signo) raise把信号发送给进程自己,相当于 raise(ing ...

  7. javascript: 字符串拼接有问题

    点击某个链接,出现js错误:Uncaught SyntaxError: Unexpected token ILLEGAL 原因: $str .= sprintf('&nbsp&nbsp ...

  8. jquery checkbox 操作

    1.jquery 获取所有选中和未选中的checkbox 未选中 var unCheckedBoxs = $("input[name='myCheckbox']").not(&qu ...

  9. Linux(centos5.0+)unison+inotify-tools触发式双向自动同步

    192.168.1.11是server1, 192.168.1.22是server2. [1]安装inotify-tools   各大linux发行版本都有inotify-tools软件包,建议通过y ...

  10. QML中的ExclusiveGroup

    Exclusive这个单词在高中应该都学过,是互斥的意思.如果你没有上过或者还没有上到高中,那你非常棒,计算机领域的大师很多都是这么起步的. ExclusiveGroup顾名思义就是互斥分组,效果很明 ...