Codeforces C1. Brain Network (easy)

最新推荐文章于 2022-10-17 09:30:35 发布

nare123

最新推荐文章于 2022-10-17 09:30:35 发布

阅读量558

点赞数

CC 4.0 BY-SA版权

分类专栏：图论

本文链接：https://blog.youkuaiyun.com/nare123/article/details/51887360

图论专栏收录该内容

7 篇文章

订阅专栏

本文介绍了一个有趣的问题：分析僵尸复杂的脑神经系统是否有效。通过给定的脑连接信息，使用图论的方法来判断该网络是否满足特定条件：任意两部分脑区能够间接或直接交换信息且移除任一连接都将破坏这一特性。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 ofn brains (much like a cow has several stomachs). They are interconnected bybrain 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.

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 from1 to n) and the number of brain connectors in the nervous system, respectively. In the nextm lines, descriptions of brain connectors follow. Every connector is given as a pair of brainsa b it connects (1 ≤ a, b ≤ n,a ≠ b).

Output

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

Examples

Input

Output

no

Input

Output

yes

大意就是给出一个图,让你判断这个图有没有环.

思路就是用矩阵存图,利用dfs或bfs来删边,如果最后还有剩余的边的话,就输出no;否则输出yes;

#include<stdio.h>
#include<queue>
using namespace std;
int mapp[1005][1005];
int vis[1005];
struct p
{
    int x,y;
};
int n,m;
queue<int>q;
void bfs(int xx)
{
    q.push(xx);
    vis[xx]=1;
    while (!q.empty())
    {
       int  fr=q.front();
       q.pop();
       for (int i=1;i<=n;i++)
       {
           if (mapp[fr][i]>0&&vis[i]==0)
           {
               vis[i]=1;
               mapp[fr][i]--;
               mapp[i][fr]--;
               q.push(i);
           }
       }
    }
}
int main()
{

   int ans=0;
   int sx,sy;
   scanf("%d%d",&n,&m);
   int flag=0;
   for (int i=1;i<=m;i++)
   {
       int a,b;
       scanf("%d%d",&a,&b);
       if (!flag)
       {
           sx=a;sy=b;
       }
       flag=1;
       mapp[a][b]++;
       mapp[b][a]++;
   }
   bfs(sx);
   for (int i=1;i<=n;i++)
   {
       for (int j=1;j<=n;j++)
       {
             if (mapp[i][j]>0)
             {
                 printf("no");
                 return 0;
             }
       }
   }
    printf("yes");
    return 0;
}