How many Tables（并差集应用）

题目描述：

Today is Ignatius' birthday. He invites a lot of friends. Now it's dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all the friends do not want to stay with strangers. 
One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, C know each other, so they can stay in one table. 

For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and D, E have to stay in the other one. So Ignatius needs 2 tables at least. 

 

InputThe input starts with an integer T(1<=T<=25) which indicate the number of test cases. Then T test cases follow. Each test case starts with two integers N and M(1<=N,M<=1000). N indicates the number of friends, the friends are marked from 1 to N. Then M lines follow. Each line consists of two integers A and B(A!=B), that means friend A and friend B know each other. There will be a blank line between two cases.
OutputFor each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks. 
Sample Input

2
5 3
1 2
2 3
4 5

5 1
2 5

Sample Output

2
4

思路：

利用并查集来找有几个集合。

代码实现 C++

#include
#include
using namespace std;
#define MAX 1010
int N,M,x,y,ans;
struct town
{
    int a;
    int b;
}t[MAX];
void Intn(int n)
{
    int i;
    ans=n;
    for(i=1;i<=n;i++)
    {
        t[i].a=i;
        t[i].b=0;
    }
}
int Find(int n)
{
    if(t[n].a==n)
        return n;
     return t[n].b=Find(t[n].a);
}
void Union(int x,int y)
{
    x=Find(x);
    y=Find(y);
    if(x==y)
        return;
        ans--;
        if(t[x].b             t[x].a=y;
    else
    {
        t[x].a=x;
        if(t[x].b>t[y].b)
            t[x].b++;
    }
}
int main()
{
    int i,O;
    cin>>O;
    while(O!=0)
    {
        if(O==0)break;
        cin>>N>>M;
        Intn(N);
        for(i=1;i<=M;i++)
        {
            cin>>x>>y;
            Union(x,y);
        }
        cout<         O--;
    }
}

总结：熟悉理解并差集的思路哦！