HDU1213 并查集模板

How Many Tables

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27984    Accepted Submission(s): 13878

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

25 31 22 34 55 12 5

 

Sample Output

24
#include "stdio.h"int tri[1005];void init(){  //初始化根节点 for(int i=1;i<1005;i++){tri[i]=i;}}int find(int x){  //找根节点 while(x!=tri[x]){x=tri[x];} return x;} bool merge(int a,int b){int c=find(a);int d=find(b);if(c!=d){   //若两者不在一个集合中 进行合并 true 合并成功 false不用合并 tri[c]=d;return true; }return false;}int main(){int kk;scanf("%d",&kk);while (kk--){init(); int n,m;scanf("%d%d",&n,&m);int x,y;for(int i=0;i<m;i++){scanf("%d%d",&x,&y);merge(x,y); }int num=0;for(int i=1;i<=n;i++){if(tri[i]==i)num++;}printf("%d\n",num);}}