How Many Tables

最新推荐文章于 2024-11-10 21:49:48 发布

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最新推荐文章于 2024-11-10 21:49:48 发布

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分类专栏：并查集

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解决Ignatius生日聚会上朋友分桌的问题，确保认识的朋友能坐在同一桌，使用并查集算法实现。

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How Many Tables

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

Total Submission(s) : 177 Accepted Submission(s) : 58

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

Sample Output

2
4

/*
title：How Many Tables
问题分类：并查集
*/
#include<stdio.h>
int tree[1010];
void init(){
   for(int i=1;i<1010;i++){
       tree[i]=i;
   }
}
int findRooot(int x){
   if(tree[x]!=x){
       x=findRooot(tree[x]);

   }
   return tree[x];
}
void merge(int x,int y){
   int fx=findRooot(x);
   int fy=findRooot(y);
   if(fx!=fy){
       tree[fy]=fx;
   }
}
int main(){
    freopen("in.txt","r",stdin);
    int zu;
    while(scanf("%d",&zu)!=EOF){
       while(zu--){
           init();
           int n,k;
           scanf("%d%d",&n,&k);
           for(int i=1;i<=k;i++){
               int a,b;
               scanf("%d%d",&a,&b);
               merge(a,b);
           }
           int count=0;
           for(int i=1;i<=n;i++){
               if(tree[i]==i){
                   count++;
               }
           }
           printf("%d\n",count);
        }

    }
    return 0;
}