hdu 1213 How Many Tables（并查集算法）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1213

How Many Tables

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13787    Accepted Submission(s): 6760

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

 

题目大意：陌生的人不会坐在一起，也就是最短路的问题，两个人之间直接有路或者是间接有路即可，找到最少需要建路的条数（即本题的桌子数）。这个题目要判断能否找到，所以需要判断哦~~

 

详见代码。

 

#include <iostream>
#include <cstdio>
using namespace std;
int father[1010];

void set(int n)
{
    for (int i=1; i<=n; i++)
        father[i]=i;
}

int find(int a)
{
    if (father[a]==a)
        return a;
    return father[a]=find(father[a]);
}

void Union(int x,int y)
{
    x=find(x);
    y=find(y);
    if (x!=y)
        father[x]=y;
}

int main ()
{
    int t,k;
    while (cin>>t)
    {

        while (t--)
        {
            k=0;
            int n,m;
            cin>>n>>m;
            set(n);
            while (m--)
            {
                int s,d;
                cin>>s>>d;
                Union(s,d);
            }
            for (int i=1; i<=n; i++)
            {
                if (father[i]==i)
                    k++;
            }
            printf ("%d\n",k);
        }
    }
    return 0;
}