hdu1213——How Many Tables（并查集）

Problem Description

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.

Input

The 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.

Output

For 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

题意：请客吃饭，希望把认识的人在一个桌子上吃饭，不认识的人在另外一桌吃饭。

思路：这是一个并查集的问题，将各自分类。

1.输入t，确定循环的次数

2.输入人数和认识的对数

3.初始化，将每个人都设置为自己认识自己init

void init(int n){

    for(int i=1;i<=n;i++)
    {
        par[i] = i;
    }
}

4.输入认识对数的具体人，将认识的人都联合在一起

int  Find(int a)
{
    if(par[a]==a)
        return a;
    else
        return Find(par[a]);
}
void Union(int a,int b)
{
    a = Find(a);
    b = Find(b);
    if(a!=b)
        par[a] = b;

}

5.输入结束后，遍历每一个节点，当它还是以自己为认识的人时，代表他需要自己做一个桌子

具体代码入下：

#include<cstdio>
using namespace std;
int par[1005];
void init(int n){

    for(int i=1;i<=n;i++)
    {
        par[i] = i;
    }
}
int  Find(int a)
{
    if(par[a]==a)
        return a;
    else
        return Find(par[a]);
}
void Union(int a,int b)
{
    a = Find(a);
    b = Find(b);
    if(a!=b)
        par[a] = b;

}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,x;
        int a,b;
        scanf("%d%d",&n,&x);
        init(n);
        for(int i=0;i<x;i++)
        {
            scanf("%d%d",&a,&b);
            Union(a,b);
        }
        int ans = 0;
        for(int i=1;i<=n;i++)
        {
            if(par[i]==i)
                ans++;
        }
        printf("%d\n",ans);

    }

    return 0;
}