Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 52483    Accepted Submission(s): 26049

 

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

 

　　典型的并查集题目，将树结构搞好以后就能得到tables

Accept

//    并查集import java.util.Scanner;public class Main{        static int T;    static int tree[];    static int n;    static int m;        public static int findFather(int node){        if(node!=tree[node]){            return tree[node]=findFather(tree[node]);        }        return node;    }        public static void main(String args[]){                Scanner reader=new Scanner(System.in);        T=reader.nextInt();        while(T>=1){            n=reader.nextInt();            m=reader.nextInt();            tree=new int[n+1];                        for(int i=1;i<=n;i++){                tree[i]=i;            }            int max=n;            for(int i=1;i<=m;i++){                int one=reader.nextInt();                int two=reader.nextInt();                int oneFather=findFather(one);                int twoFather=findFather(two);                if(oneFather!=twoFather){    //父结点不同                    if(oneFather>twoFather){    //指向更大的结点                        tree[twoFather]=oneFather;                    }else{                        tree[oneFather]=twoFather;                    }                    max--;                }            }                        System.out.println(max);            T--;        }    }    }