Bear and Friendship Condition——【并查集】
【思考】：
题目就是问你，在给出来的关系里，如果X和Y是朋友，Y和Z是朋友，X和Z是不是朋友这一个问题，有没有一点找祖先节点相不相同的感觉 ！如果有，那方向基本对了。
Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures).
There are n members, numbered 1 through n. m pairs of members are friends. Of course, a member can't be a friend with themselves.
Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z.
For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well.
Can you help Limak and check if the network is reasonable? Print "YES" or "NO" accordingly, without the quotes.
Input
The first line of the input contain two integers n and m (3 ≤ n ≤ 150 000, ) — the number of members and the number of pairs of members that are friends.
The i-th of the next m lines contains two distinct integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi). Members ai and bi are friends with each other. No pair of members will appear more than once in the input.
Output
If the given network is reasonable, print "YES" in a single line (without the quotes). Otherwise, print "NO" in a single line (without the quotes).
Examples
Input
4 3 1 3 3 4 1 4
Output
YES
Input
4 4 3 1 2 3 3 4 1 2
Output
NO
Input
10 4 4 3 5 10 8 9 1 2
Output
YES
Input
3 2 1 2 2 3
Output
NO
The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is "NO" in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not.
例子解释大意：
1 3 有边 、3 4 有边 、1 4也有边，所以是它们有关（yi）系（tui）的，
3 1 有边 、2 3 有边 、3 4 有边 、1 2 有边、但在关系里，2 4 没有相连的边，所以构不成有关系（zi ji nao bu）这一说。
ac代码：
#include
#include
#include
using namespace std;
typedef long long LL;
const int maxn=2e5+2;
int c[maxn];
int n,m;
LL v[maxn+100];//存祖先节点的边数（也就是说有多少个子节点可以到这个祖先点）
int find_set(int x)
{
if(x==c[x]) return x;
else return c[x]=find_set(c[x]);
}
int main()
{
int x,y;
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(v,0,sizeof(v));
for(int i=1;i<=n;i++)
c[i]=i;
for(int i=1;i<=m;i++)
{
scanf("%d%d",&x,&y);
int fx=find_set(x);
int fy=find_set(y);
if(fx!=fy)
{
c[fx]=fy;
}
}
for(int i=1;i<=n;i++)
v[find_set(i)]++;
LL sum=0;
for(int i=1;i<=n;i++){
if(v[i]!=0&&v[i]!=1)
{
sum+=(v[i]*(v[i]-1))/2;
}
}
if(sum==m)
printf("YES\n");
else
printf("NO\n");
}
return 0;
} 也可以用 dfs 来写，去试试别的方法吧！！！