How Many Tables-并查集

最新推荐文章于 2021-08-30 09:42:20 发布

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本文探讨了如何使用并查集数据结构解决一个关于社交聚会的问题，即确定至少需要多少张桌子来确保每个朋友都与认识的人坐在一起。通过输入不同数量的朋友和他们之间的认识关系，我们计算出所需的最少桌数。

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HDU - 1213

How Many Tables

Time Limit: 1000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

Submit Status

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

Sample Output

    
     2
4 
    
    
     题目目的很简单就是让你求有几个集合
    
    
     所以直接模板飘过
    
    
     /*
Author: 2486
Memory: 1420 KB		Time: 0 MS
Language: G++		Result: Accepted
*/
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn=1000+5;
int a,b,T,N,M,par[maxn],ranks[maxn];
void init(int sizes) {
    for(int i=1; i<=sizes; i++) {
        par[i]=i;
        ranks[i]=1;
    }
}
int find(int x) {
    return par[x]==x?x:par[x]=find(par[x]);
}
bool same(int x,int y) {
    return find(x)==find(y);
}
void unite(int x,int y) {
    x=find(x);
    y=find(y);
    if(x==y)return ;
    if(ranks[x]>ranks[y]) {
        par[y]=x;
    } else {
        par[x]=y;
        if(ranks[x]==ranks[y])ranks[x]++;
    }
}
int main() {
    scanf("%d",&T);
    while(T--) {
        scanf("%d",&N);
        scanf("%d",&M);
        init(N);
        for(int i=0; i<M; i++) {
            scanf("%d%d",&a,&b);
            unite(a,b);
        }
        int ans=0;
        for(int i=1; i<=N; i++) {
            if(par[i]==i)ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}