C - How Many Tables HDU - 1213（并查集）

最新推荐文章于 2021-12-07 22:09:42 发布

WXR1999

最新推荐文章于 2021-12-07 22:09:42 发布

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分类专栏：程序设计并查集

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本文介绍了一个解决生日派对中桌位安排问题的算法。问题的核心是如何最少地分配桌子数量，使得所有朋友都能坐在认识的人旁边。通过使用并查集数据结构，文章详细解释了如何判断哪些朋友可以坐在同一张桌子上，并提供了完整的代码实现。

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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

Sample Output

2
4

#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#define eps 1e-6
#define INF 0x3f3f3f3f

using namespace std;

int n,m,t;
int father[1010];

int Find(int x)
{

    while(x!=father[x])
        x=father[x];
    return x;
}

void Union(int a,int b)
{

    int fa=Find(a);
    int fb=Find(b);
    if(fa!=fb)
    {
        father[fa]=fb;
    }

}

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

}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {

        scanf("%d%d",&n,&m);
        init();
        for(int i=1;i<=m;i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            Union(a,b);
        }

        int ans=0;
        for(int i=1;i<=n;i++)
            if(i==Find(i))
                ans++;
        printf("%d\n",ans);
    }
    return 0;
}