Hdu 1213_hdu 1213csdn-优快云博客

本文链接：https://blog.youkuaiyun.com/wind14/article/details/54947811

本文通过一个关于朋友聚餐分桌的实际问题，介绍了并查集算法的应用。文章详细解释了如何利用并查集来判断网络的连通性，并提供了一个完整的Java实现案例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

How Many Tables

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27055 Accepted Submission(s): 13465

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

题解：本题主要考察的是并查集，讲究网络连通性

//Java代码
import java.util.Scanner;

public class Main {
    static int[] people = new int[1010];
    public static void init(){
        for(int i=1;i<people.length;i++){
            people[i] = i;
        }
    }
    public static int find(int x){//寻找根节点
        if(people[x] == x) return x;
        return find(people[x]);
    }
    public static void uSet(int a,int b){//合并
        int x = find(a);
        int y = find(b);
        if(x != y){
            people[x] = y;
        }
    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int T = in.nextInt();
        while(T-- > 0){
            init();
            int N = in.nextInt();
            int M = in.nextInt();
            for(int i=0;i<M;i++){
                int a = in.nextInt();
                int b = in.nextInt();
                uSet(a,b);
            }
            int res = 0;//结果
            for(int i=1;i<=N;i++){
                if(people[i]==i) res++;
            }
            System.out.println(res);//结果输出
            if(T!=0){
                in.nextLine();
            }
        }

    }
}