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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:
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
4 4
1 2
2 3
3 1
4 1
no
6 5
1 2
2 3
3 4
4 5
3 6
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 }
G - Brain Network (easy)(并查集水题)
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原文地址：http://www.cnblogs.com/haoabcd2010/p/5924925.html
