HDU1213 How Many Tables 并查集

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最新推荐文章于 2021-08-08 13:18:43 发布

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本文介绍了一个有趣的算法问题——如何计算Ignatius生日派对所需的最少桌数，确保熟识的朋友能坐在同一桌。通过并查集数据结构实现高效求解。

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   How Many Tables
Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 14   Accepted Submission(s) : 12

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


Author
Ignatius.L


Source
杭电ACM省赛集训队选拔赛之热身赛

来源： http://acm.hdu.edu.cn/webcontest/contest_showproblem.php?pid=1006&ojid=0&cid=10791&hide=0

#include <cstdio>
#include <algorithm>
#include <set>
using namespace std;
#define MAXN 1000*2
int Par[MAXN],Rank[MAXN];
void Init(int n)
{
    for(int i=1;i<=n;i++)
    {
        Par[i] = i;
        Rank[i] = 0;
    }
}
int  Find(int x)
{
    if(Par[x] == x) return x;
    else return Par[x] = Find(Par[x]);
}
void Unite(int x,int y)
{
    x = Find(x);y = Find(y);
    if(x == y) return;
    if(Rank[x] < Rank[y]) Par[x] = y;
    else Par[y] = x;
    if(Rank[x] == Rank[y]) Rank[x]++;
}
int main(void)
{
   // freopen("F:\\test.txt","r",stdin);
    int T;scanf("%d",&T);
    for(int t=1;t<=T;t++)
    {

        int N,M;scanf("%d %d",&N,&M);
        Init(N);
        for(int i = 1,a,b;i<=M&&scanf("%d %d",&a,&b);i++)
            Unite(a,b);
        set<int> Set;
        for(int i=1;i<=N;i++) Set.insert(Find(i));
        printf("%d\n",Set.size());
    }
}