Brain Network (easy)(并查集水题)

最新推荐文章于 2024-03-20 13:53:30 发布

转载最新推荐文章于 2024-03-20 13:53:30 发布 · 105 阅读

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原文链接：http://www.cnblogs.com/haoabcd2010/p/5924925.html

本文介绍了一个关于僵尸脑网络的问题：给定僵尸的多个大脑及连接它们的血管，判断这些大脑是否能够有效地交换思想且没有冗余连接。文章通过并查集算法解决这一问题，并提供了完整的代码实现。

G - Brain Network (easy)

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

CodeForces 690C1

Description

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.

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 a 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.

Sample Input

Input

Output

no

Input

Output

yes

//并查集水题，第一行 n , m 是点，和边个数,然后 m 边的描述，问是否都连通了且没有多余的边

 1 #include<stdio.h>
 2 int f[1001];
 3 int find(int x)
 4 {
 5     if (f[x]!=x)
 6         f[x]=find(f[x]);
 7     return f[x];
 8 }
 9 int main()
10 {
11     int n,m;
12     while (scanf("%d%d",&n,&m)!=EOF)
13     {
14         int i,j;
15         for (i=1;i<=n;i++)
16         f[i]=i;
17         
18         int a,b,temp;
19         scanf("%d%d",&a,&b);
20         temp=a;
21         f[a]=f[b];
22         int ok=1;
23         for (i=2;i<=m;i++)
24         {
25             scanf("%d%d",&a,&b);
26             int head_a=find(a);
27             int head_b=find(b);
28             if (head_a!=head_b)
29             f[head_a]=f[head_b];
30             else
31                 ok=0;
32         }
33         for (i=1;i<=n;i++)
34         {
35             if (ok==0) break;
36             if (find(i)!=find(temp))
37             {
38             ok=0;
39             break;
40             }
41         }
42         if (ok)
43         printf("yes\n");
44         else
45         printf("no\n");
46     }
47     return 0;
48 }

View Code

转载于:https://www.cnblogs.com/haoabcd2010/p/5924925.html